find all relative maxima and minima of the following function. f(x)=4x^3 - 3x^2 - 36x + 17 f(x) has a…

Explanation:

Step1: Find the derivative

Differentiate $f(x)=4x^{3}-3x^{2}-36x + 17$ using the power - rule. $f'(x)=12x^{2}-6x - 36$.

Step2: Set the derivative equal to zero

Solve $12x^{2}-6x - 36 = 0$. Divide through by 6: $2x^{2}-x - 6=0$. Factor the quadratic: $(2x + 3)(x - 2)=0$. So $x=-\frac{3}{2}$ or $x = 2$.

Step3: Find the second - derivative

Differentiate $f'(x)=12x^{2}-6x - 36$ to get $f''(x)=24x-6$.

Step4: Use the second - derivative test

For $x =-\frac{3}{2}$, $f''(-\frac{3}{2})=24\times(-\frac{3}{2})-6=-36 - 6=-42<0$. So $f(x)$ has a relative maximum at $x =-\frac{3}{2}$. $f(-\frac{3}{2})=4\times(-\frac{3}{2})^{3}-3\times(-\frac{3}{2})^{2}-36\times(-\frac{3}{2})+17=4\times(-\frac{27}{8})-3\times\frac{9}{4}+54 + 17=-\frac{27}{2}-\frac{27}{4}+54 + 17=\frac{-54 - 27}{4}+71=\frac{-81}{4}+71=\frac{-81 + 284}{4}=\frac{203}{4}=50.75$. For $x = 2$, $f''(2)=24\times2-6=48 - 6 = 42>0$. So $f(x)$ has a relative minimum at $x = 2$. $f(2)=4\times2^{3}-3\times2^{2}-36\times2+17=4\times8-3\times4-72 + 17=32-12-72 + 17=-35$.

Answer:

$f(x)$ has a relative maximum at the point $(-\frac{3}{2},50.75)$ and a relative minimum at the point $(2,-35)$. In other words, $f(x)$ has a relative maximum of $50.75$ when $x =-\frac{3}{2}$, and $f(x)$ has a relative minimum of $-35$ when $x = 2$.