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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

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10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #21 : How To Find Slope By Graphing Functions

Find the slope of the line tangent to f(x) when x is -5.

f(x)=4x^6-12x^5+3x^2

Possible Answers:

112530

Correct answer:

Explanation:

Find the slope of the line tangent to f(x) when x is -5.

f(x)=4x^6-12x^5+3x^2

To find the slope of a tangent line, first find the derivative. Then, plug in the given point. Recall the power rule of derivatives. (Multiply each term by its exponent, then subtract one from the exponent)

f(x)=4x^6-12x^5+3x^2

Becomes:

f'(x)=24x^5-60x^4+6x

Next, plug in -5

f'(-5)=24(-5)^5-60(-5)^4+6(-5)=-75000-37500-30=-112530

In doing so, we arrive at -112530, a very steep slope indeed!

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Example Question #21 : How To Find Slope By Graphing Functions

If the equation for a graph given is $\frac{3x^6+5x^2-5x+16}{x^2}$ , find the slope of a line tangent to this graph at x=5 .

Possible Answers:

1499.944

-1254.876

$-\frac{375457}{125}$

$\frac{125}{375457}$

None of the above.

Correct answer:

1499.944

Explanation:

In order to find the slope, you take derivative of the graph equation. Then by plugging in any value, you can find the slope of the graph.

In order to take the derivative of equation, the power rule must be applied, $\frac{d}{dx}x^n=nx^{n-1}$ . You must also apply the quotient rule $\frac{d}{dx}\frac{u}{v}=\frac{vu'-uv'}{v^2}$ .

Taking the derivative of the graph equation

$\frac{(18x^5+10x-5)*x^2-2x*(3x^6+5x^2-5x+16)}{x^4}$

$\frac{(18x^7+10x^3-5x^2)-(6x^7+10x^3-10x^2+32x)}{x^4}$

$\frac{(12x^7+5x^2-32x)}{x^4}$ 937,500+125-160

Plugging in x=5 , you find the slope to be $\frac{937465}{625}=1499.944$ .

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Example Question #2751 : Calculus

Take the limit

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

Possible Answers:

$\frac{2}{3}$

$\frac{8}{3}$

$\frac{5}{4}$

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 #2752 : Calculus

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 #2753 : Calculus

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:

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

$1$

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

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

$\frac{16}{5}$

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 #2754 : Calculus

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 #2755 : Calculus

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

Possible Answers:

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

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

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

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

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 #2756 : Calculus

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

Possible Answers:

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

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

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

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

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 #2757 : Calculus

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 #2758 : Calculus

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

Possible Answers:

$\frac{10000}{4}$

$\frac{9999}{4}$

$\frac{81}{4}$

$\frac{10001}{4}$

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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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #21 : How To Find Slope By Graphing Functions

Example Question #21 : How To Find Slope By Graphing Functions

Example Question #2751 : Calculus

Example Question #2752 : Calculus

Example Question #2753 : Calculus

Example Question #2754 : Calculus

Example Question #2755 : Calculus

Example Question #2756 : Calculus

Example Question #2757 : Calculus

Example Question #2758 : Calculus

All Calculus 1 Resources

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