 Talamancoeb

2021-12-31

For $f\left(x\right)=3{x}^{4}-12{x}^{3}+5$ find the following.
(A) ${f}^{\prime }\left(x\right)$
(B) The slope of the graph of f at $x=2$
(C) The equation of the tangent line at $x=2$
(D) The value(s) of x where the tangent line is horizontal
(A) f'(x) = ____
(B) At $x=2$, the slope of the graph of f is ___
(C) At $x=2$, the equation of the tangent line is y = ___
(D) The tangent line is horizontal at x =_____

(Use a comma to separate answers as needed.) Andrew Reyes

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
$\frac{d}{dx}f\left(x\right)=\frac{d}{dx}\left(3{x}^{4}-12{x}^{3}+5\right)$
${f}^{\prime }\left(x\right)=12{z}^{3}-36{x}^{2}$ Elaine Verrett

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. karton

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.
$f\left(2\right)=3\left(2{\right)}^{4}-12\left(2{\right)}^{3}+5$
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 ${f}^{\prime }\left(x\right)$ equal to 0 and solve for x to find the x values at which the tangent line is horizontal to given function.
$12{x}^{3}-36{x}^{2}=0$
$12{x}^{2}\left(x-3\right)=0$
$12{x}^{2}=0,\left(x-3\right)=0$
$x=0,\left(x-3\right)=0$
$x=0,x=3$
Step 8
Therefore, the tangent line is horizontal at x=0,3

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