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Calculus 1 : Meaning of Functions

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Example Questions

Example Question #1 : How To Find The Meaning Of Functions

Take the limit

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

Possible Answers:

$\frac{8}{3}$

$\frac{5}{4}$

$\frac{2}{3}$

Correct answer:

$\frac{5}{4}$

Explanation:

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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Example Question #2 : How To Find The Meaning Of Functions

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

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

Possible Answers:

121

Correct answer:

Explanation:

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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Example Question #3 : How To Find The Meaning Of Functions

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.

Possible Answers:

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

$1$

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

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

$\frac{16}{5}$

Correct answer:

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

Explanation:

We have the origin (0,0) and a point (x,y) 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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Example Question #4 : How To Find The Meaning Of Functions

We have the following,

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

What is c?

Possible Answers:

Correct answer:

Explanation:

First, factor the numerator of the integrand.

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

Cancel out x+6

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

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

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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Example Question #5 : How To Find The Meaning Of Functions

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

Possible Answers:

$h^{'}(x) = -cos(x)$

$h^{'}(x) = cos(x)$

$h^{'}(x) = sin(x)$

$h^{'}(x) = sin(0.5x^2)$

Correct answer:

$h^{'}(x) = sin(x)$

Explanation:

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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Example Question #6 : How To Find The Meaning Of Functions

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

Possible Answers:

$\frac{14x^6}{6}+\frac{x^5}{14}+cos(x)-cot(x)+C$

98x^6+3x^5+cos(x)+cot(x)

$\frac{14x^8}{8}+\frac{x^7}{14}+cos(x)-cot(x)+C$

$\frac{14x^8}{8}+\frac{x^7}{14}-cos(x)+cot(x)+C$

Correct answer:

$\frac{14x^8}{8}+\frac{x^7}{14}-cos(x)+cot(x)+C$

Explanation:

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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Example Question #7 : How To Find The Meaning Of Functions

Evaluate

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

Possible Answers:

55.33

66.33

Correct answer:

Explanation:

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

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

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

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

=75

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Example Question #8 : How To Find The Meaning Of Functions

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

Possible Answers:

$\frac{81}{4}$

$\frac{10001}{4}$

$\frac{9999}{4}$

$\frac{10000}{4}$

Correct answer:

$\frac{9999}{4}$

Explanation:

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

Intergation by substitution

u= x^2+1

du=2xdx

new endpoints:

u=(3)^2+1=10

u=(0)^2+1=1

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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Example Question #9 : Meaning Of Functions

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

Possible Answers:

-1

undefined

$-\infty$

Correct answer:

Explanation:

The relationship between ln(x) and x is an exponential relationship; is going to exponentially faster than ln(x) is going to $-\infty$ . One way to prove this is to write x = 1/(1/x) 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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Example Question #9 : How To Find The Meaning Of Functions

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

Possible Answers:

Essential discontinuities at x = -2 and x = -3 .

Removable discontinuity at x=2 , essential discontinuity at x=3 .

Removable discontinuities at x=2 and x=3 .

Removable discontinuities at x=-2 and x=-3 .

Removable discontinuity at x=-2 , essential discontinuity at x=-3 .

Correct answer:

Removable discontinuity at x=2 , essential discontinuity at x=3 .

Explanation:

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

Factoring both top and bottom and canceling a term x-2 tells us that this function is equal to

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

except at x=2 . This point is a removable discontinuity. x=3 is therefore an essential discontinuity where the ratio goes to $\pm \infty$ .

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Calculus 1 : Meaning of Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1 : How To Find The Meaning Of Functions

Example Question #2 : How To Find The Meaning Of Functions

Example Question #3 : How To Find The Meaning Of Functions

Example Question #4 : How To Find The Meaning Of Functions

Example Question #5 : How To Find The Meaning Of Functions

Example Question #6 : How To Find The Meaning Of Functions

Example Question #7 : How To Find The Meaning Of Functions

Example Question #8 : How To Find The Meaning Of Functions

Example Question #9 : Meaning Of Functions

Example Question #9 : How To Find The Meaning Of Functions

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