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Calculus 1 : Calculus

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

Example Questions

Example Question #332 : Other Differential Functions

Find the derivative.

2x^2-4x^3+3x-2

Possible Answers:

12x^2-4x+3

12x^2+4x+3

-12x^2+4x+3

-12x^2-4x-3

Correct answer:

-12x^2+4x+3

Explanation:

Use the product rule to find this derivative.

$\frac{d}{dx}2x^2=4x$

$\frac{d}{dx}-4x^3=12x^2$

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

Recall that the derivative of a constant is zero.

$\frac{d}{dx}-2=0$

Thus, the derivative is -12x^2+4x+3

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Example Question #336 : How To Find Differential Functions

Which of the following is an inflection point of f(x)=x^4+8x^3+18x^2+2x +5 ?

Possible Answers:

(-1,14)

(1,14)

(3,26)

(-3,14)

(-1,26)

Correct answer:

(-1,14)

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+18x^2+2x +5

f'(x)=4x^3+24x^2+36x+2

f''(x)=12x^2+48x+36

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

12x^2+48x+18=0

12(x^2+4x+3)=0

12(x+1)(x+3)=0

$x_{1,2}=-1,-3$

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

f(-1)=(-1)^4+8(-1)^3+18(-1)^2+2(-1) +5=14

f(-3)=(-3)^4+8(-3)^3+18(-3)^2+2(-3) +5=26

The two points of inflection are

(-1,14)

(-3,26)

x=-1 can be shown to be to be a point of inflection by observing the sign change at lower and higher values

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

f''(0)=12(0)^2+48(0)+36=36

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Example Question #1551 : Calculus

What is an inflection point for the function f(x)=x^3+9x^2+12 ?

Possible Answers:

(22,-1)

(3,66)

(-3,66)

(66,-3)

(66,3)

Correct answer:

(-3,66)

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+9x^2+12

f'(x)=3x^2+18x

f''(x)=6x+18

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

6x+18=0

x=-3

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

f(-3)=(-3)^3+9(-3)^2+12=66

The point of inflection is (-3,66)

It can be confirmed that x=-3 is a point of inflection due to the sign change around this point. Picking a greater and lower value x=-4,0 , observe the difference in sign of the second derivative:

f''(-4)=6(-4)+18=-6

f''(0)=6(0)+18=18

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Example Question #338 : How To Find Differential Functions

Which of the following is not an inflection point for the function $f(x)=\frac{1}{20}x^5-\frac{1}{12}x^4-x^3+15$ ?

Possible Answers:

(-2,20.1)

(4,-6.6)

(4,-19.1)

(0,15)

Correct answer:

(4,-19.1)

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:

$f(x)=\frac{1}{20}x^5-\frac{1}{12}x^4-x^3+15$

$f'(x)=\frac{1}{4}x^4-\frac{1}{3}x^3-3x^2$

f''(x)=x^3-x^2-6x

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

x^3-x^2-6x=0

x(x^2-x-6)

x(x+2)(x-3)

x=-2,0,3

These can be shown to be points of inflection by plotting f''(x)=x^3-x^2-6x and noting that it crosses the x-axis at these points; the sign of the function changes at them:

Greinflection2

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

$f(-2)=\frac{1}{20}(-2)^5-\frac{1}{12}(-2)^4-(-2)^3+15=20.1$

$f(0)=\frac{1}{20}(0)^5-\frac{1}{12}(0)^4-(0)^3+15=15$

$f(3)=\frac{1}{20}(3)^5-\frac{1}{12}(3)^4-(3)^3+15=-6.6$

The points of inflection are

(-2,20.1) , (0,15) , and (4,-6.6)

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Example Question #332 : Other Differential Functions

Find the derivative.

$\frac{2}{x^{3}}$

Possible Answers:

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

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

$\frac{6}{x^5}$

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

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

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

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Example Question #523 : Functions

Find the derivative.

$\frac{x-4}{2x}$

Possible Answers:

$\frac{1}{x^2}$

x^2

$-\frac{2}{x^2}$

$\frac{2}{x^2}$

Correct answer:

$-\frac{2}{x^2}$

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

$\frac{2x(1)-2(x-4)}{4x^2}$

$=\frac{2x-2x-8}{4x^2}$

$=-\frac{8}{4x^2}=-\frac{2}{x^2}$

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Example Question #341 : Other Differential Functions

Find the derivative.

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

Possible Answers:

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

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

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

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

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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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #332 : Other Differential Functions

Example Question #336 : How To Find Differential Functions

Example Question #1551 : Calculus

Example Question #338 : How To Find Differential Functions

Example Question #332 : Other Differential Functions

Example Question #523 : Functions

Example Question #341 : Other Differential Functions

Example Question #342 : Other Differential Functions

Example Question #524 : Differential Functions

Example Question #1553 : Calculus

All Calculus 1 Resources

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