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Calculus 1 : Calculus

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Example Questions

Example Question #43 : Meaning Of Functions

Calculate the Area of the enclosed region

Let a region, R, be bounded by the following:

y=x^3+x

y=0

x=4

Possible Answers:

Correct answer:

Explanation:

We need the top function minus the bottom function in order to determine the area enclosed between these constraints. We also need to know the bounds on the integral.

We have two functions where y=x^3+x and y=0 . Therefore, we need the first function minus the second. This is because the function zero is below the cubed function. Then, we are told that the x=4 is a constraint. That means that we are not allowed to integrate past it and that becomes our upper bound. Now we notice that the two functions with y meet at the origin, this gives us our lower bound of integration.

Recall the power rule for integration:

$\int x^n=\frac{x^{n+1}}{n+1}$

We get:

$\int_{0}^{4}((x^3+x)-(0))dx =\frac{x^4}{4}+\frac{x^2}{2}\rightarrow \frac{4^4}{4}+\frac{4^2}{2}=72$

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Example Question #50 : How To Find The Meaning Of Functions

The velocity of a car in $\textup{m/s}$ as a function of time in $\textup{s}$ is v(t)=5t^3-t^2 , where is velocity and is time.

What is the derivative of v(t) when t=4 ?

Possible Answers:

$304\textup{ m/s}^2$

$232\textup{ m/s}^2$

$232 \textup{ m/s}$

$232 \textup{ m}$

Correct answer:

$232\textup{ m/s}^2$

Explanation:

Use the Power Rule to find the derivative

v'(t)=15t^2-2t .

Substitute t=4 and get

$15(4)^2-4\cdot2=232$ .

The units of the derivative are $\textup{(units of the function)/(units of the variable) = }\textup{(m/s)/s}=\textup{m/s}^2$ .

The derivative of velocity, usually measured in meters per second (or units of length per units of time squared), is called accerelation.

The accerelation at t=4 is $232\textup{ m/s}^2$ .

Note: The Power Rule says that for a function p(x)=kx^n , $p'(x)=knx^{n-1}$ .

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Example Question #2802 : Calculus

Find the average value of $f(x)=3x^2-\frac{1}{x}$ on the interval [1,e] .

Possible Answers:

$\frac{3e^3}{e-1}$

$\frac{e}{e-1}$

$\frac{e^3}{e-1}$

$\frac{e^2}{e-1}$

$\frac{3e^2}{e-1}$

Correct answer:

$\frac{e^3}{e-1}$

Explanation:

To find the average value of a function on a closed interval [a,b], one uses the function

$\frac{1}{b-a}\int_{a}^{b}f(x)dx$ .

In this case, it is

$\frac{1}{e-1}\int_{1}^{e}3x^2-\frac{1}{x}dx$ , and using the general power rule for integral

$\int x^n \rightarrow \frac{x^{n+1}}{n+1}$

and for natural log we know,

$\int \frac{1}{x}=ln(x)$ .

Applying these rules to our function, simplfies to

$\frac{1}{e-1}(x^3-\ln x)$ evaluated from [1,e] .

This further simplies to

$\frac{1}{e-1}[(e^{3}-1)-(1-0)]=\frac{e^3}{e-1}$ .

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Example Question #1771 : Functions

Find the absolute maximum and minimum of the function f(x)=(x-2)(x+3)^2 along the interval [-4,4] .

Possible Answers:

Absolute maximum: (4,98) ; Absolute minimum (-4,-6)

Absolute maximum: (-3,0) ; Absolute minimum $\left(\frac{1}{3},-\frac{500}{27}\right)$

Absolute maximum: (4,98) ; Absolute minimum $\left(\frac{1}{3},-\frac{500}{27}\right)$

Absolute maximum: (5,192) ; Absolute minimum $\left(\frac{1}{4},-\frac{425}{24}\right)$

Absolute maximum: (3,81) ; Absolute minimum $\left(\frac{2}{3},-\frac{625}{24}\right)$

Correct answer:

Absolute maximum: (4,98) ; Absolute minimum $\left(\frac{1}{3},-\frac{500}{27}\right)$

Explanation:

Find the x-value where f'(x)=0 .

f'(x)=(x+3)^2+2(x+3)(x-2). $(x+3)^2+2(x+3)(x-2)=0\rightarrow (x+3)^2=-2(x+3)(x-2)$ .

Thus, one of the solutions is x=-3 .

Next f'(x) can be simplified to

$(x+3)=-2(x-2)\rightarrow x+3+2x-4=0\rightarrow3x-1=0\rightarrow x=\frac{1}{3}$ .

Now plug in

$x=-3, x=\frac{1}{3}, x=-4, x=4$ into f(x)=(x-2)(x+3)^2 .

$\\ f(-4)=-6, \\ \\f\left(\frac{1}{3}\right)=-\frac{500}{27},\\ \\ f(-3)=0, \\ \\f(4)=98$ .

Thus the absolute maximum is at the point (4,98) and the absolute minumum is at the point $\left(\frac{1}{3}, -\frac{500}{27}\right)$ .

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Example Question #53 : How To Find The Meaning Of Functions

At what point on the graph of $f(x)=\sqrt{x-1}$ is the tangent line parallel to the line 3x-2y=1?

Possible Answers:

(1,0)

$\left(\frac{5}{4},\frac{1}{2}\right)$

(5,2)

$\left(\frac{10}{3},\frac{1}{3}\right)$

(2,1)

Correct answer:

$\left(\frac{10}{3},\frac{1}{3}\right)$

Explanation:

First one needs to find f'(x) .

$f'(x)=\frac{1}{2}(x-1)^{-1}=\frac{1}{2\sqrt{x-1}}$ .

Then find the slope of the line 3x-2y=1 (write it in the slope-intercept to do so).

$\\-2y=-3x+1\\ \\y=\frac{3}{2}x-\frac{1}{2}$

The slope of the line is $m=\frac{3}{2}$ .

Set $f'(x)=\frac{3}{2}$ and solve for x.

$\frac{3}{2}=\frac{1}{2\sqrt{x-1}}\rightarrow\3=\frac{1}{\sqrt{x-1}}\rightarrow \frac{1}{3}=\sqrt{x-1}\rightarrow \frac{1}{9}=x-1\rightarrow x=\frac{10}{9}$ .

Plug in $x=\frac{10}{9}$ into f(x) to find the y-component of the point.

$f\left(\frac{10}{9}\right)=\frac{1}{3}$ .

Thus the point is $\left(\frac{10}{9},\frac{1}{3}\right)$ .

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Example Question #2801 : Calculus

Find the limit:

$\lim_{x \to 3} \frac{x^2+2x-15}{x^2-9}$

Possible Answers:

undefined

1/4

4/3

Correct answer:

4/3

Explanation:

If we plug directly into the function in our limit, we get an indeterminate form, 0/0, so this is an excellent candidate for using L'Hopital's rule:

$\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$

In this specific case, we get

$\lim_{x \to 3} \frac{x^2+2x-15}{x^2-9} = \lim_{x \to 3} \frac{2x+2}{2x}$

Remember, the derivative of x^n is $n*x^{n-1}$ .

If we try again to plug in 3, we get a more acceptable result of 8/6, which reduces down to 4/3.

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Example Question #2802 : Calculus

The position of a particle at a point in time is given by the function $\small \small \small x(t)=t^3-2t^2+t-12$ .

At what point in time does the velocity of the particle reach a minimum?

Possible Answers:

$\small \frac{1}{3}$

$\small 3$

$\small \frac{2}{3}$

$\small 1$

$\small \frac{3}{2}$

Correct answer:

$\small \frac{2}{3}$

Explanation:

We're given the particle's position function $\small \small x(t)=t^3-2t^2+t-11$ , we we'll first want to find the particle's velocity function. This can be found by taking the derivative of the position with respect to time:

Use the power rule to find the derivative:

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

$\small v(t)=3t^2-4t+1$

Now, to find where the velocity reaches a minimum, take the derivative one more time to find the acceleration function:

$\small a(t)=6t-4$

Find the time where the acceleration is zero, i.e. the time when the direction of the acceleration changes:

$\small 6t-4=0$

$\small \small t=\frac{2}{3}$

Note that the sign of $\small a(t)$ changes from negative to positive at this point, indicating a minimum rather than a maximum.

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Example Question #54 : How To Find The Meaning Of Functions

A huge container is used to hold all the confiscated candy at a local high school. Since everyone there loves sweets, the school is constantly confiscating candy, and dropping it into the container. At the end of the day, the container is emptied and starts empty the next day. The amount of candy in the container throughout the day is given by the following expression: $f(t) =\frac{-(t-5)^{3}}{15} +5t - 8.33$ , where t is in hours.

Assuming that there is a 10 hour school day, when is most of the candy confiscated?

Possible Answers:

Night

Unable to answer with information given

Mid-day

Morning

Constant from beginning of the day until the middle, then drops sharply

Correct answer:

Mid-day

Explanation:

Although the candy is being added throught the day, the rate in which the candy is being added is not constant throught the 10 hour school day. If the amount of candy in the container is given by the expression $f(t) =\frac{-(t-5)^{3}}{15} +5t - 8.33$ , then in order to find the rate of candy being added to the container, you must take the derivative of this expression. The derivate of a function with respect to time tells you how that expression is changing over time. If you know how quickly the contents of the container are changing, then you can know when the rate of confiscation is at its highest. Thus, the first step in this problem is to find an expression for the rate of change:

$f'(t) = \frac{-(t-5)^{2}}{5} + 5$ .

This expression represents a parabola, as shown by the t² value. At this point, there are three methods that could be used to find where the rate is the highest:

a) Theinformation ocould be extracted from the expression for the derivative. The negative t² value indicates an upside down parabola. The (t-5) indicates that the parabola was shifted 5 spaces to the right. This means that the maximum occurs at t = 5, or midday

b) The derivative could be graphed from t = 0 to t = 10 to find the time at which the maximum value is reached

Problem 4

According to the graph, this peak happens at t = 5, or midday.

c) The derivative of the derivative could be taken and set to zero in order to find the value where the derivative reaches its local maximum.

$f''(t) = -\frac{2}{5}(t-5) = 0$

t - 5=0

t = 5

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Example Question #2806 : Calculus

A toy car is thrown straight upward into the air. The equation of the position of the object is:

$y(t) = t^{2} - 5t$

How long will it take for the toy car to hit the ground?

Possible Answers:

$t = 10\: seconds$

$t = 5\: seconds$

$t = 3\: seconds$

$t = 4\: seconds$

$t = 0\: seconds$

Correct answer:

$t = 5\: seconds$

Explanation:

To find the time it takes for the toy car to hit the ground, we set the position equal to zero and we get $t = 0\: seconds$ and $t = 5\: seconds$ . At $t = 0\: seconds$ is when the toy car was launched and at $t = 5\: seconds$ is when the toy car reaches the initial position.

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Example Question #2807 : Calculus

A toy car is thrown straight upward into the air. The equation of the position of the object is:

$y(t) = t^{2} - 5t$

What is the instantaneous velocity of the toy car at t = 2 seconds?

Possible Answers:

$4\tfrac{m}{s}$

$6 \tfrac{m}{s}$

none of the answers

$1\tfrac{m}{s}$

$2 \tfrac{m}{s}$

Correct answer:

none of the answers

Explanation:

To find the velocity of the toy car, we take the derivative of the position equation. The velocity equation is

v(t) = 2t - 5

Then we insert t = 2 seconds to find the instantaneous velocity. A negative instantaneous velocity mean that the toy is slowing down.

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #43 : Meaning Of Functions

Example Question #50 : How To Find The Meaning Of Functions

Example Question #2802 : Calculus

Example Question #1771 : Functions

Example Question #53 : How To Find The Meaning Of Functions

Example Question #2801 : Calculus

Example Question #2802 : Calculus

Example Question #54 : How To Find The Meaning Of Functions

Example Question #2806 : Calculus

Example Question #2807 : Calculus

All Calculus 1 Resources

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