For f(x)=3x4−12x3+5 find the following.(A) f′(x)(B) The slope of the graph of f at x=2(C)...

Step 1
Given:
${\displaystyle f\left(x\right)=3x^4-12x^3+5}$
Step 2
A) To find ${\displaystyle f^{\prime }\left(x\right)}$
Differentiate the function with respect to x
${\displaystyle \frac{d}{dx}f\left(x\right)=\frac{d}{dx}(3x^4-12x^3+5)}$
${\displaystyle f^{\prime }\left(x\right)=12z^3-36x^2}$

Step 3
B) The slope of the ${\displaystyle f\left(x\right)}$ at ${\displaystyle x=2}$
Substitute ${\displaystyle x=2}$ in ${\displaystyle f^{\prime }\left(x\right)}$
${\displaystyle f^{\prime }\left(x\right){\mid }_{x=2}=12{\left(2\right)}^3-36{\left(2\right)}^2}$
${\displaystyle =12\left(8\right)-36\left(4\right)}$
${\displaystyle =96-144}$
${\displaystyle =-48}$
Step 4
Therefore, the slope of f at ${\displaystyle x=2}$ is -48.

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