Introductory Calculus - Pre-Calculus

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Question

Approximate the area under the following curve from to using a midpoint Riemann sum with :

$f(x)=\frac{1}{2}x^2+4$

Answer

If we want to approximate the area under a curve using n=4, that means we will be using 4 rectangles. Because the problem asks us to approximate the area from x=0 to x=4, this means we will have a rectangle between x=0 and x=1, between x=1 and x=2, between x=2 and x=3, and between x=3 and x=4. We want to use a midpoint Riemann sum, so the height of each rectangle will be the value of the function at the midpoint of each interval:

$f(x)=\frac{1}{2}x^2+4$

$f\left(\frac{1}2{}\right)=\frac{1}{2}\left(\frac{1}{2}\right)^2+4=\frac{33}{8}$

$f\left(\frac{3}{2}\right)=\frac{1}{2}\left(\frac{3}{2}\right)^2+4=\frac{41}{8}$

$f\left(\frac{5}{2}\right)=\frac{1}{2}\left(\frac{5}{2}\right)^2+4=\frac{57}{8}$

$f\left(\frac{7}{2}\right)=\frac{1}{2}\left(\frac{7}{2}\right)^2+4=\frac{81}{8}$

Now that we know the height of each rectangle, all we have to do is find its area by multiplying the height by the width, which is just 1 for each rectangle. Then we add the area of all the rectangles to find our approximation for the area under the curve from x=0 to x=4:

$A=1\left(\frac{33}{8}\right)+1\left(\frac{41}{8}\right)+1\left(\frac{57}{8}\right)+1\left(\frac{81}{8}\right)=\frac{212}{8}=\frac{53}{2}$

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Question

Using a left hand approximation and rectangles, what is the area under the curve $f(x)=\sqrt{x}$ on the interval ?

Answer

A left hand approximation requires us to slice the region into four rectangles of equal width. Since the interval on which we are slicing is and we want to create four rectangles of equal width, the width of each rectangle must be,

$\Delta x=\frac{2-0}{4}=0.5$ .

The height of each rectangular slice is given by the function value at the left edge of each rectangle. Beginning at the leftmost edge of the first rectangle on the interval, these left endpoints exist at , but not at as that is the right edge of the fourth and final rectangle on the interval.

Using the function given, $f(x)=\sqrt{x}$ , the first rectangle has height . The rest have height $f(0.5)=\sqrt{0.5}, f(1.0)=\sqrt{1}=1, f(1.5)=\sqrt{1.5}$ .

Multiply each by the width of each rectangle $\Delta x= 0.5$ to get the area of each rectangular slice. Then add the area of the slices together, and simplify to get the correct result.

$\ Area =\sum_{k=0}^{3}f(x_k)\Delta x \ \= f(0)\Delta x+f(0.5)\Delta x+f(1.0)\Delta x+f(1.5)\Delta x\ \=[f(0)+f(0.5)+f(1.0)+f(1.5)]\Delta x\$

$\x_{k} =x_{0}+k\Deltax=k\Delta x\ \ Area =0.5[0+\sqrt{0.5}+1+\sqrt{1.5}]\ \=0.5\left[\frac{1}{\sqrt{2}}+\frac{\sqrt{2}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}\right]$

$\frac{\sqrt{1}+\sqrt{2}+\sqrt{3}}{2\sqrt{2}}$

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Question

Let $f\left ( x\right )=x^{3}$

Approximate the area underneath the function on the interval divided into four sub-intervals using the midpoint height of a rectangle for each sub-interval.

Answer

The interval divided into four sub-intervals gives rectangles with vertices of the bases at

For the approximated area, we need to find the rectangle heights which values come from the midpoint of the sub-intervals, or , , , and

Because each sub-interval has width , the approximated area using rectangles is

As such, the approximated area is

units squared

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Question

Let $f\left ( x\right )=x^{2}$

Approximate the area underneath the function on the interval divided into two sub-intervals using the midpoint height of a rectangle for each sub-interval.

Answer

The interval divided into two sub-intervals gives two rectangles with vertices of the bases at

For the approximated area, we need to find the rectangle heights which values come from the midpoint of the sub-intervals, or and

Because each sub-interval has width , the approximated area using rectangles is

As such, the approximated area is

units squared

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Question

Approximate the area under the curve given by on the interval using left endpoints.

Answer

In order to approximate the area under a curve using rectangles, one must take the sum of the areas of discrete rectangles under the curve. Taking the height of each rectangle as the function evaluated at the left endpoint, we obtain the following rectangle areas:

The sum of the individual rectangles yields an overall area approximation of 100.

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Question

Approximate the area under the curve given by on the interval using right endpoints.

Answer

In order to approximate the area under a curve using rectangles, one must take the sum of the areas of discrete rectangles under the curve. Taking the height of each rectangle as the function evaluated at the right endpoint, we obtain the following rectangle areas:

The sum of the individual rectangles yields an overall area approximation of 225.

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Question

What are the discontinuities in the following function and what are their types?

$\frac{(x+1)(x-3)}{(x-3)(x-4)}$

Answer

Since the factor is in the numerator and the denominator, there is a removable discontinuity at . The function is not defined at , but function would move towards the same point for the resultant function $\frac{x+1}{x-4}$ .

Since the factor cannot be factored out, there is an infinite discontinuity at . The denominator will get very small and the numerator will move toward a fixed value.

There is no discontinuity at all at . The function simply evaluates to zero at this point.

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Question

Find the domain where the following function is continuous:

$\frac{x^4-x^3-12x^2}{x+3}$

Answer

The function in the numerator factors to:

so if we cancel the x+3 in the numerator and denominator we have the same function but it is continuous. The $\frac{x+3}{x+3}$ gives us a hole at x=-3 so our function is not continuous at x=-3.

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Question

Determine if the function $y=x^\frac{2}{3}$ is continuous at using limits.

Answer

In order to determine if a function is continuous at a point three things must happen.

Taking the limit from the lefthand side of the function towards a specific point exists.

Taking the limit from the righthand side of the function towards a specific point exists.

The limits from 1) and 2) are equal and equal the value of the original function at the specific point in question.

In our case,

$\lim_{x->0^{-}}=0$

$\lim_{x->0^{+}}=0$

Because all of these conditions are met, the function is continuous at 0.

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Question

Determine if is continuous on all points of its domain.

Answer

First, find that at any point where , .

Then find that

$\lim_{x->b^{+}} e^{x}=e^{b}$ and $\lim_{x->b^{-}} e^{x}=e^{b}$ .

As these are all equal, it can be determined that the function is continuous on all points of its domain.

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Question

Let $f(x)=\frac{x^{2}-9}{x-3}$ . Determine if the function is continuous using limits.

Answer

As approaches , the function $f(x)=\frac{x^{2}-9}{x-3}$ approaches $\frac{3^{2}-9}{3-3}=\frac{0}{0}$ , which is undefined. However, if we factor , we get:

$f(x)=\frac{x^{2}-9}{x-3}$

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

The factors in the numerator and denominator cancel out, leaving .

Therefore, our function is continuous at all values of from $(-\infty, \infty)$ .

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Question

Determine the location of the points of inflection for the following function:

$f(x)=\frac{1}{12}x^4-\frac{1}{2}x^3-5x^2+4x+17$

Answer

The points of inflection of a function are the points at which its concavity changes. The concavity of a function is described by its second derivative, which will be equal to zero at the inflection points, so we'll start by finding the first derivative of the function:

$f(x)=\frac{1}{12}x^4-\frac{1}{2}x^3-5x^2+4x+17$

$f'(x)=\frac{1}{3}x^3-\frac{3}{2}x^2-10x+4$

Next we'll take the derivative one more time to get the second derivative of the original function:

$f'(x)=\frac{1}{3}x^3-\frac{3}{2}x^2-10x+4$

Now that we have the second derivative of the function, we can set it equal to 0 and solve for the points of inflection:

$x^2-3x-10=0\rightarrow (x-5)(x+2)=0\rightarrow x=-2,x=5$

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Question

Which of the following is an -coordinate of an inflection point of the graph of the following function?

$f(x)=\frac{1}{12}x^4-2x^3+16x^2-19x+36$

Answer

The inflection points of a function are the points where the concavity changes, either from opening upwards to opening downwards or vice versa. The inflection points occur at the x-values where the second derivative is either zero or undefined. That means we need to find our second derivative.

We start by using the Power Rule to find the first derivative.

$f'(x)=\frac{1}{3}x^3-6x^2+32x-19$

Taking the derivative once more gives the second derivative.

We then set this derivative equal to zero and solve.

This factors nicely.

Therefore our second derivative is zero when

8 is the only one of these two amongst our choices and is therefore our answer.

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Question

Determine the points of inflection of the following function:

$f(x)=\frac{1}{12}x^4+\frac{1}{3}x^3-\frac{15}{2}x^2+17$

Answer

The points of inflection of a function are those at which its second derivative is equal to 0. First we find the second derivative of the function, then we set it equal to 0 and solve for the inflection points:

$f(x)=\frac{1}{12}x^4+\frac{1}{3}x^3-\frac{15}{2}x^2+17$

$f'(x)=\frac{1}{3}x^3+x^2-15x$

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Question

Find the inflection points of the following function:

$f(x)=\frac{1}{6}x^4-x^3-28x^2+9$

Answer

The points of inflection of a function are those at which its second derivative is equal to 0. First we find the second derivative of the function, then we set it equal to 0 and solve for the inflection points:

$f(x)=\frac{1}{6}x^4-x^3-28x^2+9$

$f'(x)=\frac{2}{3}x^3-3x^2-56x$

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Question

Find the points of inflection of the following function:

$f(x)=\frac{1}{4}x^4-5x^3+\frac{63}{2}x^2-21$

Answer

The points of inflection of a function are those at which its second derivative is equal to 0. First we find the second derivative of the function, then we set it equal to 0 and solve for the inflection points:

$f(x)=\frac{1}{4}x^4-5x^3+\frac{63}{2}x^2-21$

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Question

Determine the points of inflection, if any, of the following function:

Answer

The points of inflection of a function are those at which its concavity changes. The concavity of a function is described by its second derivative, and when the second derivative is 0 a point of inflection occurs. We find the second derivative of the function and then set it equal to 0 to solve for the inflection points:

$6x-10=0\rightarrow 6x=10\rightarrow x=\frac{5}{3}$

So the function has only one point of inflection at x=5/3.

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Question

Determine the x-coordinate of the inflection point of the function .

Answer

The point of inflection exists where the second derivative is zero.

, and we set this equal to zero.

$x=\frac{8}{6}=\frac{4}{3}$

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Question

Find the point of inflection of the function .

Answer

To find the x-coordinate of the point of inflection, we set the second derivative of the function equal to zero.

$x=\frac{6}{12}=\frac{1}{2}$ .

To find the y-coordinate of the point, we plug the x-coordinate back into the original function.

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

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

$y=\frac{7}{2}$

The point is then $\left(\frac{1}{2},\frac{7}{2}\right)$ .

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Question

Find the x-coordinates of all points of inflection of the function .

Answer

We set the second derivative of the function equal to zero to find the x-coordinates of any points of inflection.

, and the quadratic formula yields

$x=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}$ .

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