How to find the meaning of functions - CLEP Calculus

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Question

If $h(x)=\int_{0}^{x}sin(u)du$ , find $h^{'}(x)$

Answer

Taking the derivative of an integral yields the original function, but because we have a different variable in the integration limits, the variable switches

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Question

Evaluate $\int(14x^7 +\frac{x^6}{(2)}+sin(x)-csc^2(x))dx$

Answer

using integration identities: $\int{x^n dx}= \frac{x^{n+1}}{n+1} and \int{sin(x) dx}= - cos(x) and \int{csc^2(x) dx}= -cot (x)$

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Question

Evaluate the limit:

$\lim_{x \to 2} \frac{x^3-8}{x^2-4}$

Answer

First, factor both the numerator and denominator:

$\lim_{x \to 2} \frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)}$

Cancel:

$\lim_{x \to 2} \frac{(x^2+2x+4)}{(x+2)}$

Plugging in 2 gives the final answer, 3.

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Question

Take the limit

$\lim_{x \to -2} \frac{x+\sqrt{x+6}}{x+2}$

Answer

First, multiply the numerator and denominator by $x-\sqrt{x+6}$ and it turns into

$\lim_{x \to -2} \frac{(x+\sqrt{x+6})}{(x+2)}\frac{(x-\sqrt{x+6})}{(x-\sqrt{x+6})}$

$\lim_{x \to -2} \frac{x^2-x-6}{(x+2)(x-\sqrt{x+6})}$

Factor the numerator and then cancel out the 'x+2'

$\lim_{x \to -2} \frac{(x+2)(x-3)}{(x+2)(x-\sqrt{x+6})}$

$\lim_{x \to -2} \frac{(x-3)}{(x-\sqrt{x+6})}$

After taking the limit, the answer is $\frac{5}{4}$

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Question

If this limit is true, then what is the value of 'a'?

$\lim_{x \to 3} \frac{x^4-a}{x^2-\sqrt{a}}=18$

Answer

Factor the numerator

$\lim_{x \to 3} \frac{(x^2-\sqrt{a})(x^2+\sqrt{a})}{x^2-\sqrt{a}}=18$

Cancel the $x^2-\sqrt{a}$ , plug in the limit and then solve for 'a'

$\lim_{x \to 3} x^2+\sqrt{a}=18$

$3^2+\sqrt{a}=18$

$a=81$

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Question

We have a line described as $y=-\frac{1}{2}x+2$ . Find the minimum distance between the origin and a point on that line.

Answer

We have the origin and a point located on the line. That point represents the minimum distance to the orgin. Apply the distance formula to these two points,

$d=\sqrt{(x-0)^2+(y-0)^2}=\sqrt{x^2+y^2}$

Plug in the line equation, take the derivative, set it equal to zero, and solve for x.

$d=\sqrt{x^2+(-\frac{1}{2}x+2)^2}=\sqrt{\frac{5}{4}x^2-2x+4}$ ${d}'=\frac{\frac{5}{2}x-2}{2\sqrt{\frac{5}{4}x^2-2x+4}}=0$

$x=\frac{4}{5}$

Use this value to find

$y=-\frac{1}{2}\left ( \frac{4}{5} \right )+2=\frac{8}{5}$

So we have the point $(\frac{4}{5},\frac{8}{5})$ , which is closest to the origin. We can now find its distance from that origin.

$d=\sqrt{\left (\frac{4}{5} \right )^2+\left ( \frac{8}{5} \right )^2}$

$d=\sqrt{\frac{16}{5}}=\frac{4}{\sqrt{5}}$

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Question

We have the following,

$\int_{-5}^{5}\frac{x^2+(6+c)x+6c}{x+6}dx=-60$

What is c?

Answer

First, factor the numerator of the integrand.

$\int_{-5}^{5}\frac{(x+6)(x+c)}{x+6}dx$

Cancel out

$\int_{-5}^{5}\frac{(x+6)(x+c)}{x+6}dx=-60$

$\int_{-5}^{5}(x+c)dx=-60$

Perform the integral and then solve for

$\frac{1}{2}(5)^2+5c-\left ( \frac{1}{2}(-5)^2+(-5)c \right )=-60$

$10c=-60$

$c=-6$

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Question

Evaluate

$\int_{1}^{4}(4x^2+2x-8)dx$

Answer

$\int_{1}^{4}(4x^2+2x-8)dx$

$\frac{4x^3}{3} + \frac{2x^2}{2} -8x$

$\frac{4x^3}{3} + x^2 -8x$ evaluate at $\left [ 1,4 \right ]$

$\left(\frac{4(4)^3}{3} + (4)^2 -8(4)\right) - \left(\frac{4(1)^3}{3} + (1)^2 -8(1)\right)$

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Question

Evaluate $\int_{0}^{3}(2x(x^2+1)^3)dx$

Answer

$\int_{0}^{3}(2x(x^2+1)^3)dx$

Intergation by substitution

new endpoints:

New Equation:

$\int_{1}^{10}(u^3)du$

$\frac{u^4}{4}$ at $\left [ 1,10 \right ]$

$= \frac{(10)^4}{4} - \frac{(1)^4}{4}$

$= \frac{(10000)}{4} - \frac{(1)}{4}$

$= \frac{9999}{4}$

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Question

What is $\lim_{x \to 0^+} ~~x \cdot \ln(x)$ ?

Answer

The relationship between and x is an exponential relationship; is going to exponentially faster than is going to $-\infty$ . One way to prove this is to write and use L'Hôpital's rule:

$\lim_{x \to 0^+} \frac{\ln x}{1/x} = \lim_{x \to 0^+} \frac{\ln'x}{(1/x)'} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} \frac{-x^2}{x} = -0 = 0$

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Question

Where is $\frac{4x^3 - 21 x + 10}{x^2 - 5x + 6}$ discontinuous? Are those discontinuities removable?

Answer

The rational function $\frac{4x^3 - 21 x + 10}{x^2 - 5x + 6}$ has a denominator with two roots, and . These are discontinuities.

Factoring both top and bottom and canceling a term tells us that this function is equal to

$\frac{(2x - 1) (2x + 5)}{x - 3}$

except at . This point is a removable discontinuity. is therefore an essential discontinuity where the ratio goes to $\pm \infty$ .

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Question

The derivative of is written $y^{(n)}(x)$ . The factorial of is written .

Find the function defined on such that $f^{(n)}(0) = n!$

Answer

The MacLaurin series is:

$\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} ~x^n$ ,

which in this case is

$\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + \dots$ .

This is a standard geometric series, and can be informally solved for by writing:

$f(x) = 1 + x \cdot f(x)$

Hence,

$f(x) = \frac{1}{1 - x}$ .

Since this is informal, we must check that the derivatives indeed still match. Fortunately, they do:

$f^{(n)}(x) = \frac{n!}{(1 - x)^{n + 1}} ; ~~ f^{(n)}(0) = n!$

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Question

A farmer has 500 feet of fencing. She wants to enclose the maximum area possible, so she plans to use the side of another building, which is 300 feet long, as one side of her enclosure. Which of the following sets of equations should she use to solve her problem?

Answer

Equation 1: $\small A=lw$

Equation 2: $\small 500=l+2w$

$\small A$ is the area of the enclosure, $\small l$ is the length of the enclosure, and $\small w$ is the width of the enclosure.

Equation 1 represents the quantity the farmer wants to maximize -- the enclosed area. Equation 2 represents the constraint on the area -- the perimeter of the enclosed area. We know that the perimeter must equal 500, but we do not know how long or wide the enclosure should be.

Equation 1: $\small A=s^2$

Equation 2: $\small 500=4s$

$\small A$ is the area of the field and $\small s$ is the length of each side.

This set of equations incorrectly assumes that the enclosure must be a square.

Equation 1: $\small A=lw$

Equation 2: $\small 500=2l+2w$

$\small A$ is the area of the enclosure, $\small l$ is the length of the enclosure, and $\small w$ is the width of the enclosure.

This set of equations does not take into account the building's being used as a side of the enclosure. No fencing wil be used for that side.

Equation 1: $\small A=300w$

Equation 2: $\small 500=300+2w$

$\small A$ is the area of the enclosure and $\small w$ is the width of the enclosure.

This set of equations assumes that the enclosure must be the length of the building used as the fourth side. While the length of the building does determine the maximum length of the enclosure, the length of the enclosure can be less. It is also incorrect in that it accounts for the length of only three sides.

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Question

A farmer has 500 feet of fencing. She wants to enclose the maximum area possible, so she plans to use the side of another building, which is 300 feet long, as one side of her enclosure. What equation will you use to find the maximum area the farmer can enclose?

Answer

The farmer wants to maximize her enclosed area using 500 feet of fencing and one side of a 300-foot-long building. We can model this situation using the following set of equations:

Equation 1: $\small A=lw$

Equation 2: $\small 500=l+2w$

Equation 1 expresses what we want to maximize and Equation 2 expresses the constraint on that function. $\small A$ is the area, $\small l$ is the length of the enclosure, and $\small w$ is the width of the enclosure. The length is included in the constraint equation only once because the building will serve as the fourth side of the enclosure; we don't need to use any of our fencing along that side.

Now we can solve Equation 2 for $\small l$

$\small l=500-2w$

Which permits us to re-write Equation 1 in terms of only one variable

$\small A=(500-2w)w$

$\small A=500w-2w^2$

The Extreme Value Theorem tells us that any closed and bounded continuous function must attain a maximum within the interval on which it is defined. The width must be at least zero (it can't be meaningfully negative) and at most 250 feet (which would use up all 500 feet of our fencing while forcing the length to be zero). So our function is closed, bounded, and continuous, and all we need to do is find its first derivative, which is $\small A'=500-4w$ .

$\small A'=250-4w$ is incorrect because it is the derivative of $\small A=250w-2w^2$ , which was likely arrived at after incorrectly setting up the initial constraint equation as $\small 500=2l+2w$ .

$\small A'=500w-2w^2$ fails to differentiate $\small A$ .

$\small A'=500-2w$ is an incorrect application of the Power Rule, which fails to multiply the coefficient of $\small w^2$ by the exponent.

The extremes of our function will occur at the end points and at any critical points. We already know that the end points of our function give an area of zero (because either the length or the width is zero at those points) and so all we have to do is set the first derivative equal to zero and solve for $\small w$

$\small A'=500-4w$

$\small 0=500-4w$

$\small 4w=500$

$\small w=125$

Now we know that the maximum area occurs when the width is 125 feet, so we just solve for $\small l$

$\small 500=l+2(125)$

$\small 250=l$

Thus, the maximum area is achieved by making the length 250 feet and the width 125 feet.

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Question

A farmer has 500 feet of fencing. She wants to enclose the maximum area possible, so she plans to use the side of another building, which is 300 feet long, as one side of her enclosure. What is the maximum area the farmer can enclose?

Answer

The farmer wants to maximize her enclosed area using 500 feet of fencing and one side of a 300-foot-long building. We can model this situation using the following set of equations:

Equation 1: $\small A=lw$

Equation 2: $\small 500=l+2w$

Equation 1 expresses what we want to maximize and Equation 2 expresses the constraint on that function. $\small A$ is the area, $\small l$ is the length of the enclosure, and $\small w$ is the width of the enclosure. The length is included in the constraint equation only once because the building will serve as the fourth side of the enclosure; we don't need to use any of our fencing along that side.

Now we can solve Equation 2 for $\small l$

$\small l=500-2w$

Which permits us to re-write Equation 1 in terms of only one variable

$\small A=(500-2w)w$

$\small A=500w-2w^2$

The Extreme Value Theorem tells us that any closed and bounded continuous function must attain a maximum within the interval on which it is defined. The width must be at least zero (it can't be meaningfully negative) and at most 250 feet (which would use up all 500 feet of our fencing while forcing the length to be zero). So our function is closed, bounded, and continuous, and all we need to do is find its first derivative, which is $\small A'=500-4w$ .

The extremes of our function will occur at the end points and at any critical points. We already know that the end points of our function give an area of zero (because either the length or the width is zero at those points) and so all we have to do is set the first derivative equal to zero and solve for $\small w$

$\small A'=500-4w$

$\small 0=500-4w$

$\small 4w=500$

$\small w=125$

Now we know that the maximum area occurs when the width is 125 feet, so we just solve for $\small l$

$\small 500=l+2(125)$

$\small 250=l$

Thus, the maximum area is achieved by making the length 250 feet and the width 125 feet, or 31,250 ft2.

11,718.75 ft2 is incorrect because it relies on an incorrect constraint equation.

500 ft2 is incorrect because it inappropriately substitutes the given numbers for area and length in the formula $\small A=lw$ .

37,500 ft2 is incorrect because it relies on the assumption that the rectangle must be square.

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Question

Which of the following functions is NOT differentiable at $\small x=0$ ?

Answer

A function is differentiable over an interval if its graph is continuous and has no corners or vertical tangent lines. A visual inspection of the graph can reveal whether a function is continuous, however it is safer to take the limit at the point in question.

Here, all four functions are continuous, and graphing reveals a potential vertical tangent line only for $\small f(x)=x^\frac{1}{3}$ at $\small x=0$ . If we take the limit

$\small \lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

$\small \lim_{h\to0}\frac{(0+h)^\frac{1}{3}-(0)^\frac{1}{3}}{h}$

$\small \lim_{h\to0}\frac{h^\frac{1}{3}}{h}$

$\small \small \lim_{h\to0}\frac{1}{h^\frac{2}{3}}$

As $\small h$ goes to zero the denominator becomes small and the fraction grows without bound. For example,

when $\small h=1$ , then $\small \frac{1}{1^\frac{2}{3}}=1$

when $\small h=0.1$ , then $\small \frac{1}{0.1^\frac{2}{3}}=\frac{1}{0.000001}=1,000,000$

and when $\small \small h=0.01$ , then $\small \frac{1}{0.01^\frac{2}{3}}=\frac{1}{0.000000000001}=1,000,000,000,000$

In other words, as $\small x$ gets closer and closer to zero, $\small y$ becomes increasingly large, until the function is finally undefined at $\small x=0$ .

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Question

Compute the value of the following limit:

$\lim_{x\rightarrow 0}\frac{3x^3-4x^2}{11x^2-5x^3}$

Answer

Plugging x=0 into the equation, we can see that we get a result of 0/0 . Whenever we evaluate a limit and get a result of 0/0, remember that this means we can apply L’Hopital’s rule to the equation. This means we take the derivative of the numerator and denominator separately, replacing the numerator and denominator with their respective derivatives, and evaluate the limit:

$\lim_{x\rightarrow 0}\frac{9x^2-8x}{22x-15x^2}$

After we plug in x=0 once again, we see that the result is still 0/0, so we must take the derivative of the numerator and denominator once again, and then plug in x=0 to see if we obtain an appropriate value for the limit:

$\lim_{x\rightarrow 0}\frac{18x-8}{22-30x}=\frac{18(0)-8}{22-30(0)}=-\frac{8}{22}=-\frac{4}{11}$

We no longer get a result of 0/0, and can see the value of the limit is -4/11.

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Question

For the following function, determine the absolute extrema on the interval :

Answer

Any extrema, local or global, will occur when the slope of the function becomes 0, so we must first find our critical points to identify where the slope of the function is 0:

So our critical points are at x=0, x=6, and x= -4. We can see that the last critical point, x= -4, is outside of the interval specified in the question, so this point can be thrown out. Now that we know two critical points within our interval, we can check the value of the function at these points, as well as the endpoints of the interval, to see which is the absolute maximum and which is the absolute minimum:

Here we can see that our absolute maximum of 8209 occurs at x=6, while our absolute minimum of 1 occurs at x=0.

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Question

For the function below, find the inflection points and use them to determine the intervals on which the function is concave up and concave down:

Answer

To find the inflection points and concavity of any function, we must first find its second derivative. If we set the equation for the second derivative equal to 0, we can then solve for the x values of our function’s inflection points:

Because the function is a polynomial, we know it is continuous everywhere, and therefore we know our two inflection points are valid. To find the intervals on which the funciton is concave up and concave down, we simply plug in a value below our first inflection points, a value between our inflection points, and a value above our second inflection point into the equation for f ’’(x) and see if we get a negative or positive value. Negative indicates the function is concave down on the interval, while positive indicates the function is concave up on the interval.

Plugging in arbitrary values below, between, and above our inflection points, we can see the second derivative is negative (the funciton is concave down) on (0,6), and positive (the function is concave up) on (-∞,0) and (6,∞).

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Question

Suppose $y=\frac{3x^3+3x-2}{2x^3-2x+3}$ . What is the limit as approaches infinity?

Answer

Using the method of substitution will give an indeterminate form of $\frac{\infty}{\infty}$ .

Therefore, it is possible to use L'Hopital to determine the limit.

Write the formula for L'Hopital.

$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}$

Take the derivative of the numerator and denominator of $\lim_{x\rightarrow \infty}\frac{3x^3+3x-2}{2x^3-2x+3}$ .

$\lim_{x\rightarrow \infty}\frac{9x^2+3}{6x^2-2} =\frac{\infty}{\infty}$

Since substitution will yield an indeterminate form, apply L'Hopital again.

$\lim_{x\rightarrow \infty}\frac{18x}{12x} = \frac{18}{12} = \frac{3}{2}$

The limit is $\frac{3}{2}$ as $x\rightarrow\infty$ .

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