let $h(x)=\frac{1}{x^{3}}-\frac{2}{x^{2}}$. $h(-1)=square$
let $h(x)=\frac{1}{x^{3}}-\frac{2}{x^{2}}$. $h(-1)=square$
Answer
Explanation:
Step1: Find the derivative of (h(x))
Use the power - rule ((x^n)^\prime=nx^{n - 1}). For (h(x)=\frac{1}{3}x^{3}-\frac{1}{2}x^{2}), then (h^\prime(x)=x^{2}-x).
Step2: Evaluate (h^\prime(-1))
Substitute (x = - 1) into (h^\prime(x)). So (h^\prime(-1)=(-1)^{2}-(-1)).
Step3: Simplify the expression
(h^\prime(-1)=1 + 1=2).
Answer:
2