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Interactive-12Adv-Concavity & points of inflection

Interactive-12Adv-Concavity & points of inflection

Assessment

Interactive Video

•

Mathematics

•

12th Grade

•

Practice Problem

•

Hard

Created by

ASHTON TSELEPY

FREE Resource

6 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

For what values of x is the curve y=4x3−2x2+x+1y=4x^3-2x^2+x+1 concave upwards?

x<16x<\frac{1}{6}

x<−1/6x<-1/6

x>−1/6x>-1/6

x>16x>\frac{1}{6}

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Find the domain over which the curve y=4x3−2x2+x+1y=4x^3-2x^2+x+1 is concave downwards.

(−∞,16)(-∞,\frac{1}{6})

(73,∞)\left(\frac{7}{3},∞\right)

(−∞,−16)(-∞,-\frac{1}{6})

(−73,∞)\left(-\frac{7}{3},∞\right)

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What condition must be met for a point of inflection to occur?

The first derivative is zero.

The second derivative is greater than zero.

The second derivative is zero and the concavity changes.

The second derivative is less than zero.

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What conditions are required for a horizontal point of inflection?

f''(x) = 0 and f'(x) > 0

f''(x) = 0 and f'(x) = 0

f''(x) > 0 and f'(x) = 0

f''(x) < 0 and f'(x) = 0

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What are the x-values for which the second derivative of the function f(x)=3x5−10x3+7f(x)=3x^5-10x^3+7 is equal to zero?

x=0,x=1x=0,x=1

x=−1,x=1x=-1,x=1

x=0,x=−1,x=1x=0,x=-1,x=1

x=0,x=−1x=0,x=-1

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What condition must be met for a point to be classified as a horizontal point of inflection?

The first derivative is zero and the second derivative changes sign.

The second derivative is zero and the second derivative changes sign.

The first derivative is zero and the second derivative does not change sign.

The second derivative is zero and the second derivative does not change sign.

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