find the absolute maximum and absolute minimum values of the function, if they exist, over the indicated…

Explanation:

Step1: Find the derivative

$f'(x)=3x^{2}+2x - 5=(3x + 5)(x - 1)$

Step2: Find critical points

Set $f'(x)=0$, then $(3x + 5)(x - 1)=0$. So $x = 1$ or $x=-\frac{5}{3}$. But we are considering the interval $(0,\infty)$, so we discard $x =-\frac{5}{3}$.

Step3: Evaluate the function at the critical - point and analyze the behavior

Evaluate $f(x)$ at $x = 1$: $f(1)=1^{3}+1^{2}-5\times1 + 4=1 + 1-5 + 4=1$. As $x\to\infty$, $y = f(x)=x^{3}+x^{2}-5x + 4\to\infty$ since the leading - term $x^{3}$ has a positive coefficient and the degree is odd.

Answer:

C. No absolute maximum; absolute minimum is 1