for f(x)=3x^4 - 4x^3 + 5 find the following. (a) f(x) (b) the slope of the graph of f at x = - 2 (c) the…

Explanation:

Step1: Find the derivative of f(x)

Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, for $f(x)=3x^{4}-4x^{3}+5$, we have $f^{\prime}(x)=3\times4x^{3}-4\times3x^{2}=12x^{3}-12x^{2}$.

Step2: Find the slope at x = - 2

Substitute $x=-2$ into $f^{\prime}(x)$. So $f^{\prime}(-2)=12(-2)^{3}-12(-2)^{2}=12\times(-8)-12\times4=-96 - 48=-144$.

Step3: Find the equation of the tangent line at x = - 2

First, find $f(-2)=3(-2)^{4}-4(-2)^{3}+5=3\times16+4\times8 + 5=48 + 32+5=85$. The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(-2,85)$ and $m=-144$. So $y - 85=-144(x + 2)$, which simplifies to $y=-144x-288 + 85=-144x-203$.

Step4: Find where the tangent line is horizontal

Set $f^{\prime}(x)=0$, so $12x^{3}-12x^{2}=12x^{2}(x - 1)=0$. Solving $12x^{2}(x - 1)=0$ gives $x = 0$ or $x = 1$.

Answer:

(A) $12x^{3}-12x^{2}$ (B) $-144$ (C) $-144x-203$ (D) $0,1$