Relationship between the increasing and decreasing behavior of ƒ and the sign of ƒ' - AP Calculus AB

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Question

On which interval(s) is the function increasing and on which interval(s) is it decreasing?

Answer

First we must find out critical points by setting equal to zero.

Then we reverse foil to get

$0=(3x-2)(x+4)\Rightarrow 3x-2=0 \and\ x+4=0$

$x=\frac{2}{3}, x=-4$

These are our critical points. We now must test values on each interval defined by our critical points in order to determine the sign of on each interval.

Our intervals to check are

$(-\infty,-4),(-4,\frac{2}{3}),(\frac{2}{3},\infty)$

We have many choices, but let's choose $x=-5, \ x=0 ,\ x=1$

$f'(-5)=3\cdot (-5^2)+10\cdot(-5)-8=17$

$f'(0)=3\cdot0^2 +10\cdot0-8=-8$

$f'(1)=3\cdot (1^2)+10\cdot (1)-8=5$

Thus on our first interval is positive meaning is increasing.

On our second interval is negative meaning is decreasing.

And on our third interval is positive meaning is increasing.

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Question

At what point does $4x^2+\frac{2}{3}x$ shift from decreasing to increasing?

Answer

To find out where it shifts from decreasing to increasing, we need to look at the first derivative. The shift will happen where the first derivative goes from a negative value to a positive value.

To find the first derivative for this problem, we can use the power rule. The power rule states that we lower the exponent of each of the variables by one and multiply by that original exponent.

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(24x^{2-1})+(1\frac{2}{3}x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(24x^{1})+(1\frac{2}{3}x^{0})$

Remember that anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(8x)+(1*\frac{2}{3}(1))$

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=8x+\frac{2}{3}$

Can this equation be negative? Yes. Does it shift from negative to positive? Yes. Therefore, it will shift from negative to positive at the point that $0=8x+\frac{2}{3}$ .

$0=8x+\frac{2}{3}$

$-\frac{2}{3}=8x$

$\frac{-\frac{2}{3}}{8}=x$

$-\frac{1}{12}=x$

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Question

Find the derivative of g(t) and tell whether g(t) is increasing or decreasing on the interval \[5,6\].

Answer

Find the derivative of g(t) and tell whether g(t) is increasing or decreasing on the interval \[5,6\]

First, find the derivative by decreasing each exponent by 1 and multiplying the coefficient by that number.

Next, plug in our two endpoints of our interval to see what the sign of g'(t) is.

Now, clearly these are both negative, and every point between them will be negative. This means that function g(t) is decreasing on this interval.

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Question

What does the sign of the first derivative tell us about whether a function is increasing or decreasing?

Answer

What does the sign of the first derivative tell us about whether a function is increasing or decreasing?

The first derivative test is used to tell whether a function is increasing or decreasing at a certain point or interval.

To use this test, first find the derivative of your function. Then, plug in the values for the point(s) and see what sign you get on your values.

If the value of your first derivative is negative, then your function is decreasing. If the value of your first derivative is positive, your original function is increasing. If your first derivative is 0, then you have a point of inflection in your original function.

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Question

Use the first derivative test to tell whether f(c) is increasing or decreasing when c=24.

Answer

Use the first derivative test to tell whether f(c) is increasing or decreasing when c=24

Begin by finding the first derivative of f(c)

Next, plug in 24 for c and find the sign of our first derivative.

Now, our first derivative is positive, so our original function must be increasing.

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Question

Tell whether f is increasing or decreasing when . How do you know?

Answer

Tell whether f is increasing or decreasing when . How do you know?

To test for increasing/decreasing, we need to find the first derivative.

In this case, we can use the power rule to do all our differentiation.

Power rule:

$\frac{d}{dx}x^n=(n)x^{n-1}$

We will use this on each term in order to find our first and then second derivative.

For each term, we will decrease the exponent by 1, and then multiply by the original exponent.

Now, we need to find the sign of f'(-12). This will tell us if it is increasing or decreasing.

So, we get

So,

f(x) is decreasing, because

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Question

Determine the intervals on which the function is increasing:

$f=\frac{x^5}{5}-\frac{26}{3}x^3+25x$

Answer

To determine the intervals on which the function is increasing, we must determine the intervals on which the function's first derivative is positive.

The first derivative of the function is equal to

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Next, we must find the critical values, at which the first derivative is equal to zero:

$c=\pm1, \pm5$

Using the critical values, we now create intervals on which to evaluate the sign of the first derivative:

$(-\infty, -5), (-5, -1), (-1, 1), (1, 5), (5, \infty)$

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is positive, on the second interval, the first derivative is negative, on the third interval, the first derivative is positive, on the fourth interval, the first derivative is negative, and on the fifth interval, the first derivative is positive. Therefore, the function is increasing on the intervals $(-\infty, -5)\cup(-1,1)\cup(5, \infty)$ .

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Question

Determine the intervals on which f is decreasing, for its entire domain:

Answer

To determine the intervals on which the function is decreasing, we must determine the intervals on which the function's first derivative is negative.

The first derivative of the function is equal to

and was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Next, we must find the critical values, at which the first derivative is equal to zero:

Using the critical values, we now create intervals over which to evaluate the sign of the first derivative:

$(-\infty, -6), (-6,0), (0, \infty)$

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

On the first interval, the first derivative is negative, while on the remaining intervals, the first derivative is positive. Thus, the function is decreasing on the first interval, $(-\infty, -6)$ .

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Question

Determine the intervals on which f is decreasing:

$f=x^5+\frac{x^3}{3}$

Answer

To determine the intervals on which the function is decreasing, we must determine the intervals on which the function's first derivative is negative.

The first derivative of the function is equal to

and was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Next, we must find the critical values, at which the first derivative is equal to zero:

The square root of a negative number is not valid for a critical value, so we only have one critical value for this function.

Using the critical value, we now create intervals over which to evaluate the sign of the first derivative:

$(-\infty, 0), (0, \infty)$

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

On the first interval, the first derivative is positive, and on the second interval, the first derivative is positive as well. The function is therefore never decreasing because the first derivative is never negative.

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Question

Determine the intervals on which f is increasing:

$f=\cos(5x), 0\leq x \leq \frac{2\pi}{5}$

Answer

To determine the intervals on which the function is increasing, we must determine the intervals on which the function's first derivative is positive.

The first derivative of the function is equal to

$f'=-5\sin(5x)$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Next, we must find the critical values, at which the first derivative is equal to zero:

$-5\sin(5x)=0$

$c=0, \frac{\pi}{5}, \frac{2\pi}{5}$

Note that we stopped writing critical values based on the given interval in the problem statement.

Using the critical values, we now create intervals over which to evaluate the sign of the first derivative:

$(0, \frac{\pi}{5}), (\frac{\pi}{5}, \frac{2\pi}{5})$

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is negative, while on the second interval, the first derivative is positive. Thus, the function is increasing on the second interval, $(\frac{\pi}{5}, \frac{2\pi}{5})$ .

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Question

Find the local maxima for the following function:

$f= \frac{x^3}{3}+x^2+x$

Answer

To determine the local maxima of the function, we must determine the points at which the function's first derivative changes from positive to negative.

The first derivative of the function is

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Next, we must find the critical values, at which the first derivative is equal to zero:

Using the critical value as an endpoint, we now create intervals over which to evaluate the sign of the first derivative:

$(-\infty, -1), (-1, \infty)$

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is positive, and on the second interval, the first derivative is positive as well. There is no sign change of the first derivative, thus there are no local maxima (the function is always increasing).

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Question

Determine the local minima of the following function:

$f=\frac{(x-1)^4}{4}+27x$

Answer

To determine the local minima of the function, we must determine the points at which the function's first derivative changes from negative to positive.

The first derivative of the function is

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Next, we must find the critical values, at which the first derivative is equal to zero:

Using the critical value as an endpoint, we now create intervals over which to evaluate the sign of the first derivative:

$(-\infty, -2), (-2, \infty)$

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is negative, and on the second interval, the first derivative is positive. There is a sign change from negative to positive of the first derivative at , thus a local minima exists here.

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Question

Determine the local minima of the following function:

Answer

To determine the local minima of the function, we must determine the points at which the function's first derivative changes from negative to positive.

The first derivative of the function is

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Next, we must find the critical values, at which the first derivative is equal to zero:

Using the critical value as an endpoint, we now create intervals over which to evaluate the sign of the first derivative:

$(-\infty, 0), (0, \infty)$

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is negative, and on the second interval, the first derivative is also negative. The first derivative never changed in sign from negative to positive, thus the function has no local minima.

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Question

Determine the intervals on which the function is increasing:

Answer

To determine the intervals on which the function is increasing, we must determine the intervals on which the function's first derivative is positive.

The first derivative of the function is equal to

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Next, we must find the critical values, at which the first derivative is equal to zero:

Using the critical values, we now create intervals on which to evaluate the sign of the first derivative:

$(-\infty, -3), (-3, -1), (-1, \infty)$

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is positive, on the second interval, the first derivative is negative, and on the third interval, the first derivative is positive. Thus, the function is increasing on the first and third intervals, $(-\infty, -3)\cup(-1, \infty)$ .

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Question

Determine the intervals on which the function is decreasing:

$f(x)=\frac{x^3}{3}+2x^2+x$

Answer

To determine the intervals on which the function is decreasing, we must determine the intervals on which the function's first derivative is negative.

The first derivative of the function is equal to

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Next, we must find the critical values, at which the first derivative is equal to zero:

$x+2=\pm \sqrt{3}$

$c=-2\pm \sqrt{3}$

(Note that the method of completing the square was shown for solving for the critical values. One could use the quadratic formula as well.)

Using the critical values, we now create intervals over which to evaluate the sign of the first derivative:

$(-\infty, -2-\sqrt{3}), (-2-\sqrt{3}, -2+\sqrt{3}),(-2+\sqrt{3}, \infty)$

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is positive, on the second interval, the first derivative is negative. and on the third interval, the first derivative is positive. Thus, the interval on which the first derivative is negative is the interval where the function is decreasing, $(-2-\sqrt{3}, -2+\sqrt{3})$ .

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Question

Let $f(x) = \frac{1}{7}x^{7} - \frac{6}{5} x^{5} + \frac{5}{3}x^{3} - 30x$ .

A relative minimum of the graph of can be located at:

Answer

At a relative minimum of the graph , it will hold that and .

First, find . Using the sum rule,

$f'(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1}{7}x^{7} - \frac{6}{5} x^{5} + \frac{5}{3}x^{3} - 30x \right )$

$f'(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1}{7}x^{7} \right )- \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{6}{5} x^{5} \right )+ \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{5}{3}x^{3} \right )- \frac{\mathrm{d} }{\mathrm{d} x}\left ( 30x \right )$

Differentiate the individual terms:

$f'(x) = \frac{1}{7} \cdot 7 x^{6} - \frac{6}{5} \cdot 5 x^{4} + \frac{5}{3} \cdot 3 x^{2} - 30x$

$f'(x) = x^{6} - 6 x^{4} +5 x^{2} - 30$

Set this equal to 0:

$x^{6} - 6 x^{4} +5 x^{2} - 30 = 0$

$(x^{6} - 6 x^{4} )+(5 x^{2} - 30) = 0$

$x^{4} (x^{2} - 6 )+ 5 (x^{2} - 6 ) = 0$

$(x^{4}+ 5 ) (x^{2} - 6 ) = 0$

Either:

$x^{4}+5 = 0$ , in which case, $x^{4} = -5$ ; this equation has no real solutions.

$x^{2}- 6=0$ has two real solutions, $-\sqrt {6}$ and $\sqrt {6}$ .

Now take the second derivative, again using the sum rule:

$f'(x) = x^{6} - 6 x^{4} +5 x^{2} - 30$

$f''(x) = \frac{\mathrm{d} }{\mathrm{d} x}(x^{6} - 6 x^{4} +5 x^{2} - 30)$

$f''(x) = \frac{\mathrm{d} }{\mathrm{d} x}(x^{6} )- \frac{\mathrm{d} }{\mathrm{d} x} ( 6 x^{4} )+\frac{\mathrm{d} }{\mathrm{d} x}( 5 x^{2} )- \frac{\mathrm{d} }{\mathrm{d} x}( 30)$

$f''(x) =6 x^{5} - 6 ( 4 x^{3} )+ 5( 2x )- 0$

$f''(x) =6 x^{5} -2 4 x^{3} +10x$

Substitute $-\sqrt {6}$ for :

$f''(-\sqrt {6}) =6 (-\sqrt {6} )^{5} -2 4 (-\sqrt {6})^{3} +10(-\sqrt {6})$

$f''(-\sqrt {6}) =- 216\sqrt {6}+144 \sqrt {6} -10\sqrt {6}$

$f''(-\sqrt {6}) =-82\sqrt {6} < 0$

Therefore, has a relative maximum at $x =- \sqrt {6}$ .

Now. substitute $\sqrt {6}$ for :

$f''( \sqrt {6}) =6 ( \sqrt {6} )^{5} -2 4 ( \sqrt {6})^{3} +10( \sqrt {6})$

$f''(-\sqrt {6}) = 216\sqrt {6}-144 \sqrt {6} +10\sqrt {6}$

$f''(-\sqrt {6}) =82\sqrt {6} >0$

Therefore, has its only relative minimum at $x = \sqrt {6}$ .

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Question

Find the intervals on which the function is decreasing:

$f=\frac{x^4}{4}-{\frac{x^3}{3}}-2x^2+4x$

Answer

To determine the intervals on which the function is decreasing, we must determine the intervals on which the function's first derivative is negative.

The first derivative of the function is equal to

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Next, we must find the critical values, at which the first derivative is equal to zero:

Note that factoring by grouping was used to find the critical values.

Using the critical values, we now create intervals over which to evaluate the sign of the first derivative:

$(-\infty, -2), (-2, 1), (1, 2), (2, \infty)$

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is negative, on the second interval, the first derivative is positive, on the third interval, the first derivative is negative, and on the fourth interval, the first derivative is positive. The intervals on which the function is decreasing are the intervals on which the first derivative was negative, $(-\infty, -2)\cup(1, 2)$ .

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Question

Determine the intervals on which the following function is increasing:

$f=\frac{x^3}{3}-x$

Answer

To determine the intervals on which the function is increasing, we must determine the intervals on which the function's first derivative is positive.

The first derivative of the function is equal to

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Next, we must find the critical values, at which the first derivative is equal to zero:

Using the critical values, we now create intervals over which to evaluate the sign of the first derivative:

$(-\infty, -1), (-1, 1), (1, \infty)$

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is positive, on the second interval, the first derivative is negative, and on the third interval, the first derivative is positive. The function is therefore increasing on two intervals, $(-\infty, -1)\cup(1, \infty)$ , on which we just showed the first derivative was positive.

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Question

Find the intervals on which the function is decreasing:

$f=\frac{x^3}{3}+4x^2-20x$

Answer

To determine the intervals on which the function is decreasing, we must determine the intervals on which the function's first derivative is negative.

The first derivative of the function is equal to

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Next, we must find the critical values, at which the first derivative is equal to zero:

Using the critical values, we now create intervals on which to evaluate the sign of the first derivative:

$(-\infty, -10), (-10, 2), (2, \infty)$

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is positive, on the second interval, the first derivative is negative, and on the third interval, the first derivative is positive. Thus, we know that the function is decreasing on the second interval, .

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Question

Determine the local maxima of the function:

$f=2\cos(3x), 0\leq x \leq \pi$

Answer

To determine the values at which the function has a local maximum, we must determine the values at which the sign of the first derivative changes from positive to negative.

The first derivative of the function is equal to

$f'=-6\sin(3x)$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Next, we must find the critical values, at which the first derivative is equal to zero:

$-6\sin(3x)=0$

$c=0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi$

Using the critical values, we now create intervals on which to evaluate the sign of the first derivative:

$(0, \frac{\pi}{3}), (\frac{\pi}{3}, \frac{2\pi}{3}), (\frac{2\pi}{3}, \pi)$

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the first derivative, we find that on the first interval, the first derivative is negative, on the second interval, the first derivative is positive, and on the third interval, the first derivative is negative. The first derivative changes from positive to negative at $x=\frac{2\pi}{3}$ , thus there is exists a local maximum.

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