for f(x)=x^4 - 8x^3 + 4 find the following. (a) f(x) (b) the slope of the graph of f at x = - 3 (c) the…

for f(x)=x^4 - 8x^3 + 4 find the following. (a) f(x) (b) the slope of the graph of f at x = - 3 (c) the equation of the tangent line at x = - 3 (d) the value(s) of x where the tangent line is horizontal (a) f(x)= (b) at x = - 3, the slope of the graph of f is. (c) at x = - 3, the equation of the tangent line is y =. (d) the tangent line is horizontal at x =. (use a comma to separate answers as needed.)

for f(x)=x^4 - 8x^3 + 4 find the following. (a) f(x) (b) the slope of the graph of f at x = - 3 (c) the equation of the tangent line at x = - 3 (d) the value(s) of x where the tangent line is horizontal (a) f(x)= (b) at x = - 3, the slope of the graph of f is. (c) at x = - 3, the equation of the tangent line is y =. (d) the tangent line is horizontal at x =. (use a comma to separate answers as needed.)

Answer

Explanation:

Step1: Find the derivative of f(x)

Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, for $f(x)=x^{4}-8x^{3}+4$, we have $f'(x)=4x^{3}-24x^{2}$.

Step2: Find the slope at x = - 3

Substitute $x=-3$ into $f'(x)$. So $f'(-3)=4(-3)^{3}-24(-3)^{2}=4\times(-27)-24\times9=-108 - 216=-324$.

Step3: Find the point on the function at x = - 3

First, find $f(-3)=(-3)^{4}-8(-3)^{3}+4=81 + 216+4=301$. The point is $(-3,301)$. Using the point - slope form of a line $y - y_{1}=m(x - x_{1})$ with $m=-324$, $x_{1}=-3$ and $y_{1}=301$, we get $y - 301=-324(x + 3)$. Expand it: $y-301=-324x-972$, so $y=-324x - 671$.

Step4: Find where the tangent line is horizontal

Set $f'(x)=0$, so $4x^{3}-24x^{2}=4x^{2}(x - 6)=0$. Then $x = 0$ or $x = 6$.

Answer:

(A) $4x^{3}-24x^{2}$ (B) $-324$ (C) $-324x - 671$ (D) $0,6$