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Calculus 1 : How to find differential functions

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Example Questions

Example Question #341 : How To Find Differential Functions

Find the derivative.

$\frac{4}{x^3}$

Possible Answers:

$-\frac{6}{x^4}$

$-\frac{4}{x^4}$

$-\frac{6}{x^3}$

$-\frac{4}{x^3}$

Correct answer:

$-\frac{6}{x^4}$

Explanation:

Use the quotient rule to find this derivative.

Remember that the quotient rule is:

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

Apply this to our problem to get

$\frac{d}{dx}\frac{4}{x^3}=\frac{x^3(0)-2(3x^2)}{x^6}=\frac{-6x^2}{x^6}=\frac{-6}{x^4}$

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Example Question #342 : Other Differential Functions

Find the derivative at x=2.

$\frac{1}{x^4}$

Possible Answers:

$-\frac{1}{16}$

$\frac{1}{4}$

$\frac{1}{16}$

$-\frac{1}{8}$

Correct answer:

$-\frac{1}{8}$

Explanation:

First, find the derivative using the quotient rule.

Remember that the quotient rule is:

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

Apply this to our problem to get

$\frac{d}{dx}\frac{1}{x^4}=\frac{x^4(0)-1(4x^3)}{x^8}=\frac{-4x^3}{x^8}$

$=-\frac{4}{x^5}$

Now, substitute 2 for x.

$-\frac{4}{2^5}=-\frac{4}{32}=-\frac{1}{8}$

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Example Question #524 : Differential Functions

Which of the following is an inflection point for the function f(x)=x^4+8x^3-5x-13 ?

Possible Answers:

(4,13)

(-4,-13)

(0,-249)

(0,-13)

(0,-4)

Correct answer:

(0,-13)

Explanation:

The points of inflection of a function occur where the second derivative of the funtion is equal to zero.

Find this second derivative by taking the derivative of the function twice:

f(x)=x^4+8x^3-5x-13

f'(x)=4x^3+24x^2-5

f''(x)=12x^2+48x

Set the second derivative to zero and find the values that satisfy the equation:

12x^2+48x=0

12x(x+4)=0

x=0,-4

These can be shown to be points of inflection by the change in sign of the second derivative at points just below and after these points:

For x=0

12(0.5)((0.5)+4)=27

12(-0.5)((-0.5)+4)=-21

For x=-4

12(-3)((-3)+4)=-36

12(-5)((-5)+4)=60

Now, plug these values back in to the original function to find the values of the function that match to them:

f(0)=0^4+8(0)^3-5(0)-13=-13

f(-4)=(-4)^4+8(-4)^3-5(-4)-13=-249

The points of inflection are

(0,-13)

(-4,-249)

x=0 can be shown to be to be a point of inflection by observing the sign change at lower and higher values on the second derivative.

f''(-1)=12(-1)^2+48(-1)=-36

f''(1)=12(1)^2+48(1)=60

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Example Question #1553 : Calculus

Which of the following is an inflection point for the function f(x)=x^3+21x^2+10 ?

Possible Answers:

(2,102)

(-3,172)

(172,-3)

(696,-7)

(-7,696)

Correct answer:

(-7,696)

Explanation:

The points of inflection of a function occur where the second derivative of the funtion is equal to zero.

Find this second derivative by taking the derivative of the function twice:

f(x)=x^3+21x^2+10

f'(x)=3x^2+42x

f'(x)=6x+42

Set the second derivative to zero and find the values that satisfy the equation:

6x+42=0

x=-7

It can be shown that this is a point of inflection by the change in sign of the second derivative with points before and after this value

6(-6)+42=6

6(-8)+42=-6

Now, plug this value back in to the original function to find the value of the function that matches:

f(x)=(-7)^3+21(-7)^2+10=696

The point of inflection is (-7,696)

It can be confirmed that x=-7 is a point of inflection due to the sign change around this point. Picking a greater and lower value x=-8,0 , observe the difference in sign of the second derivative:

f''(-8)=6(-8)+42=-6

f''(0)=6(0)+42=42

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Example Question #532 : Differential Functions

Which of the following is not a point of inflection for the function $f(x)=\frac{1}{3}x^3+2\pi cos(x)$ ?

Possible Answers:

$(\pi, 4.05)$

(-2.48,-10.04)

(1.18,2.94)

$(-\pi,-16.62)$

Correct answer:

$(\pi, 4.05)$

Explanation:

The points of inflection of a function occur where the second derivative of the funtion is equal to zero.

Find this second derivative by taking the derivative of the function twice:

$f(x)=\frac{1}{3}x^3+2\pi cos(x)$

To take these derivatives, make use of the following rules:

d[sin(u)]=cos(u)du

d[cos(u)]=-sin(u)du

$f'(x)=x^2-2\pi sin(x)$

$f''(x)=2x-2\pi cos(x)$

Set the second derivative to zero and find the values that satisfy the equation:

$2x-2\pi cos(x)=0$

$x=-\pi,-2.48,1.18$

These can be shown to be points inflection by observing how the signs on the plot of $f''(x)=2x-2\pi cos(x)$ ; note how the function crosses the x-axis at these values

Greinflectionplot

Now, plug these values back in to the original function to find the values of the function that match to them:

$f(-\pi)=\frac{1}{3}(-\pi)^3+2\pi cos(-\pi)=-16.62$

$f(-2.48)=\frac{1}{3}(-2.48)^3+2\pi cos(-2.48)=-10.04$

$f(1.18)=\frac{1}{3}(1.18)^3+2\pi cos(1.18)=2.94$

The points of inflection are $(-\pi,-16.62)$ , (-2.48,-10.04) , (1.18,2.94)

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Example Question #1561 : Calculus

Find the derivative at x=3 .

$7x^{2}+x-6$

Possible Answers:

Correct answer:

Explanation:

First, find the derivative using the power rule:

Remember that the power rule is:

$(x^n)'=nx^{n-1}$

Apply this to our problem to get

14x+1

Now, substitute for .

14(3)+1=43

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Example Question #531 : Differential Functions

Find the derivative at x=6 .

2x+x^2-9

Possible Answers:

Correct answer:

Explanation:

First, use the power rule to find the derivative.

Remember that the power rule is:

$(x^n)'=nx^{n-1}$

Apply this to our problem to get

2+x

Now, substitute for .

2+6=8

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Example Question #535 : Differential Functions

Find the derivative at x=0 .

x^4+2x^3-3

Possible Answers:

$\frac{1}{2}$

Correct answer:

Explanation:

Find the derivative using the power rule.

Remember that the power rule is:

$(x^n)'=nx^{n-1}$

Apply this to our problem to get

4x^3+6x^2

Now, substitute for .

4(0)^3+6(0)^2

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Example Question #536 : Differential Functions

Find the derivative at x=1 .

5x^3+2x^2-x

Possible Answers:

Correct answer:

Explanation:

First, find the derivative using the power rule.

Remember that the power rule is:

$(x^n)'=nx^{n-1}$

Apply this to our problem to get

15x^2+4x-1

Now, substitute for .

15(1)^2+4(1)-1=15+4-1=18

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Example Question #531 : Differential Functions

Find the derivative.

7x^3+x^2+3x

Possible Answers:

7x^2+x+3

21x+2

7x^3-x^2-3x

21x^2+2x+3

Correct answer:

21x^2+2x+3

Explanation:

Use the power rule to find this derivative.

Remember that the power rule is:

$(x^n)'=nx^{n-1}$

Apply this to our problem to get

$\frac{d}{dx}7x^3=21x^2$

$\frac{d}{dx}x^2=2x$

$\frac{d}{dx}3x=3$

Thus, the derivative is 21x^2+2x+3 .

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Calculus 1 : How to find differential functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #341 : How To Find Differential Functions

Example Question #342 : Other Differential Functions

Example Question #524 : Differential Functions

Example Question #1553 : Calculus

Example Question #532 : Differential Functions

Example Question #1561 : Calculus

Example Question #531 : Differential Functions

Example Question #535 : Differential Functions

Example Question #536 : Differential Functions

Example Question #531 : Differential Functions

All Calculus 1 Resources

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