Talamancoeb

2021-12-31

For

(A)

(B) The slope of the graph of f at

(C) The equation of the tangent line at

(D) The value(s) of x where the tangent line is horizontal

(A) f'(x) = ____

(B) At

(C) At

(D) The tangent line is horizontal at x =_____

(Use a comma to separate answers as needed.)

Andrew Reyes

Beginner2022-01-01Added 24 answers

Step 1

Given:

$f\left(x\right)=3{x}^{4}-12{x}^{3}+5$

Step 2

A) To find${f}^{\prime}\left(x\right)$

Differentiate the function with respect to x

$\frac{d}{dx}f\left(x\right)=\frac{d}{dx}(3{x}^{4}-12{x}^{3}+5)$

$f}^{\prime}\left(x\right)=12{z}^{3}-36{x}^{2$

Given:

Step 2

A) To find

Differentiate the function with respect to x

Elaine Verrett

Beginner2022-01-02Added 41 answers

Step 3

B) The slope of the$f\left(x\right)$ at $x=2$

Substitute$x=2$ in ${f}^{\prime}\left(x\right)$

$f}^{\prime}\left(x\right){\mid}_{x=2}=12{\left(2\right)}^{3}-36{\left(2\right)}^{2$

$=12\left(8\right)-36\left(4\right)$

$=96-144$

$=-48$

Step 4

Therefore, the slope of f at$x=2$ is -48.

B) The slope of the

Substitute

Step 4

Therefore, the slope of f at

karton

Expert2022-01-04Added 556 answers

Step 5

NSK

C) The equation of the tangent line at x = 2

First find the y coordinates by substituting x=2 in the given function.

y=48-96+5

y=-43

Step 6

Therefore, the equation of the tangent line passing through (2,-43) with slope -48, is

y-(-43)=-48(x-(-2))

y+43=-48x+96

y=-48x+96-43

y=-48x+53

Step 7

D)

The value(s) of x where the tangent line is horizontal

Set

Step 8

Therefore, the tangent line is horizontal at x=0,3

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