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Calculus 1 : Calculus

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Example Questions

Example Question #22 : How To Find The Meaning Of Functions

Evaluate:

${}\lim_{t \rightarrow 10} \frac{e^{t^2 - 100}-1}{t-10}$

Possible Answers:

{}10

{}20

{}40

Limit Does Not Exist

Correct answer:

{}20

Explanation:

$\\ \lim_{t \rightarrow 10} \frac{e^{t^2 - 100}-1}{t-10}$

If we plug in 10 we get, $\ \frac{1-1}{0}=\frac{0}{0}$ .

So we can use L'Hospital's Rule.

L'Hospital's Rule states that if

$\\ \lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\frac{\pm\infty}{\pm\infty}$ or $\\ \lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\frac{0}{0}$ , where is any real number then,

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

Set $f(t)=e^{t^2-100}-1$ , and g(t)=t-10 .

Then find f'(t) , and g'(t) .

$f'(t)=2t\cdot e^{t^2-100}$

g'(t)=1

We can now rewrite our limit as:

$\\ \\ \lim_{t \rightarrow 10} \frac{2t\cdot e^{t^2-100}}{1} = \frac{2\cdot 10\cdot1}{1} =20$

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Example Question #23 : How To Find The Meaning Of Functions

Find $\small \lim_{x\rightarrow 0}f(x)$ if .

Possible Answers:

$\small \infty$

$\small 0$

Does not exist

$\small 4$

$\small 1$

Correct answer:

$\small 4$

Explanation:

When we sub in $\small 0$ for $\small x$ , we get $\small 0/0$ , so we should try to factor the expression.

We can factor out an $\small x$ in the numerator to get,

$\small x(x+4)/x$ .

After canceling out the $\small x$ , the expression reduces to,

(x+4) ,

subbing in zero for $\small x$ we get $\small 4$ . That is our limit.

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Example Question #31 : How To Find The Meaning Of Functions

Evaluate:

$\lim_{x\rightarrow 10} \frac{x^3-30x^2+300x-1000}{x-10}$

Possible Answers:

The limit does not exist.

Correct answer:

Explanation:

In order to evaluate the limit, lets factor the numerator.

$\lim_{x\rightarrow 10} \frac{(x-10)^3}{x-10}$

Now we can simplify this expression to

$\lim_{x\rightarrow 10} (x-10)^2$ .

Now we plug in 10.

$\lim_{x\rightarrow 10} (x-10)^2=(10-10)^2=0$

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Example Question #1755 : Functions

Evaluate:

$\lim_{x\rightarrow\pi} \frac{\sin^2(x)}{\cos\left(\frac{x}{2}\right)}$

Possible Answers:

$\pi$

$\frac{\pi}{2}$

The limit does not exist.

Correct answer:

Explanation:

If we plug in $\pi$ , we get

$\lim_{x\rightarrow\pi} \frac{\sin^2(x)}{\cos(\frac{x}{2})}=\frac{\sin^2(\pi)}{\cos{(\frac{\pi}{2})}}=\frac{0}{0}$

Since we get $\frac{0}{0}$ , we can use L'Hopitals rule.

L'Hopitals rule is if we have one of the following cases

$\lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\frac{0}{0}$

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

where a is any real number, then

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

After applying L'Hopitals rule, we get

$\lim_{x\rightarrow\pi} \frac{2\sin(x)\cos(x)}{-\frac{1}{2}\sin(\frac{x}{2})}$

Now if we plug in $\pi$ , we get

$\lim_{x\rightarrow\pi} \frac{2\sin(x)\cos(x)}{-\frac{1}{2}\sin(\frac{x}{2})}=\frac{2\sin(\pi)\cos(\pi)}{-\frac{1}{2}\sin(\frac{\pi}{2})}=\frac{2\cdot 0\cdot 1}{-\frac{1}{2}\cdot1}=0$

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Example Question #1756 : Functions

Let C(t) represent the growth rate of the city of Tucson at a year where t=0 represents the year 2000 . Give a practical interpretation of $\int_0^5C(t) dt$ .

Possible Answers:

The total increase in people in Tucson from 2000 to 2005 .

The rate of change in the number of people in 2005 .

The average rate of change of people between 2000 and 2005 .

The average number of people between 2000 and 2005 .

The number of cats in people in 2005 .

Correct answer:

The total increase in people in Tucson from 2000 to 2005 .

Explanation:

The integral of the rate of change of people gives the change in the number of people over the time interval. However, it does not tell you how many people are present because it contains no information on how many people were present initially. (Think that if you integrate f'(x) you get . Here denotes the number of people added and denotes the initial number of people.)

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Example Question #32 : How To Find The Meaning Of Functions

Find the critical point(s) of $f(x)=4x^{2}+9x-6$ .

Possible Answers:

$c=-\frac{9}{8}$ and $c=\frac{9}{8}$

$c=-\frac{9}{8}$

$c=\frac{9}{8}$

$c=-\frac{8}{9}$ and $c=\frac{9}{8}$

$c=-\frac{9}{8}$ and $c=\frac{8}{9}$

Correct answer:

$c=-\frac{9}{8}$

Explanation:

To find the critical point(s) of a function f(x) , take its derivative f'(x) , set it equal to , and solve for .

Given $f(x)=4x^{2}+9x-6$ , use the power rule

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ to find the derivative. Thus the derivative becomes, f'(x)=8x+9=0 .

Since 8x+9=0 :

8x=-9

$x=-\frac{9}{8}$

The critical point is $c=-\frac{9}{8}$ .

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Example Question #33 : How To Find The Meaning Of Functions

Find the critical point(s) of $f(x)=9x^{2}-12x+7$ .

Possible Answers:

$c=\frac{3}{2}$ and $c=-\frac{3}{2}$

$c=-\frac{2}{3}$

$c=\frac{2}{3}$ and $c=-\frac{2}{3}$

$c=\frac{3}{2}$ and $c=-\frac{2}{3}$

$c=\frac{2}{3}$

Correct answer:

$c=\frac{2}{3}$

Explanation:

To find the critical point(s) of a function f(x) , take its derivative f'(x) , set it equal to , and solve for .

Given $f(x)=9x^{2}-12x+7$ , use the power rule $f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ to find the derivative.

Thus the derivative becomes, f'(x)=18x-12=0 .

Since 18x-12=0 :

18x=12

$x=\frac{12}{18}=\frac{2}{3}$

The critical point is $c=\frac{2}{3}$ .

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Example Question #1752 : Functions

Which of the following is not a function?

Possible Answers:

$f(x)=\left\{\begin{matrix}3x+2\textup{ when } x> 7 & \\3x-2\textup{ when } x\leq 7 \end{matrix}\right.$

$f(x)=\left\{\begin{matrix}3x+2\textup{ when } x>7 & \\3x-2\textup{ when } x< 7 \end{matrix}\right.$

$f(x)=\left\{\begin{matrix}3x+2\textup{ when } x\geq 7 & \\3x-2\textup{ when } x\leq 7 \end{matrix}\right.$

f(x)=3x+2

Correct answer:

$f(x)=\left\{\begin{matrix}3x+2\textup{ when } x\geq 7 & \\3x-2\textup{ when } x\leq 7 \end{matrix}\right.$

Explanation:

A function is defined when each value of x yields a single value of y (ordered pairs). The expression

$f(x)=\left\{\begin{matrix}3x+2\textup{ when } x\geq 7 & \\3x-2\textup{ when } x\leq 7 \end{matrix}\right.$

is not a function because there are two values of f(x) for which a single value of x (7) creates. f(7) can be either 3(7)+2=19 or 3(7)-2=21, and is therfore not a function.

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Example Question #32 : How To Find The Meaning Of Functions

Evaluate:

$\lim_{x\rightarrow1} \frac{x^2-2x+1}{x-1}$

Possible Answers:

$\infty$

The limit does not exist

Correct answer:

Explanation:

To evaluate the limit, we can factor the numerator

$\lim_{x\rightarrow1} \frac{x^2-2x+1}{x-1}=\lim_{x\rightarrow1} \frac{(x-1)^2}{x-1}$

Now we simplify and evaluate.

$\lim_{x\rightarrow1} \frac{(x-1)^2}{x-1}=\lim_{x\rightarrow1} x-1=1-1=0$

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Example Question #33 : Meaning Of Functions

Evaluate:

$\lim_{x\rightarrow 1} \frac{x^3-6x^2+11x-6}{x-1}$

Possible Answers:

$\infty$

Correct answer:

Explanation:

In order to evaluate the limit, we need to factor the numerator

$\lim_{x\rightarrow 1} \frac{x^3-6x^2+11x-6}{x-1}=\lim_{x\rightarrow 1}\frac{(x-1)(x-2)(x-3)}{x-1}$

Now we can simplify the expression to

$\lim_{x\rightarrow1} (x-2)(x-3)$ .

Now we can plug in 1 to get

$\lim_{x\rightarrow1} (x-2)(x-3)=(1-2)(1-3)=-1\cdot -2=2$ .

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #22 : How To Find The Meaning Of Functions

Example Question #23 : How To Find The Meaning Of Functions

Example Question #31 : How To Find The Meaning Of Functions

Example Question #1755 : Functions

Example Question #1756 : Functions

Example Question #32 : How To Find The Meaning Of Functions

Example Question #33 : How To Find The Meaning Of Functions

Example Question #1752 : Functions

Example Question #32 : How To Find The Meaning Of Functions

Example Question #33 : Meaning Of Functions

All Calculus 1 Resources

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