Question

in parts (a)-(c), find the indicated quantity for y = f(x)=8 - x².
(a) $\frac{f(-1)-f(-2)}{(-1)-(-2)}=square$
(b) $\frac{f(-2 + h)-f(-2)}{h}=square$
(c) $lim_{h
ightarrow0}\frac{f(-2 + h)-f(-2)}{h}=square$

Explanation:

Step1: Find f(-1)

Substitute x = - 1 into f(x)=8 - x², so f(-1)=8-(-1)²=8 - 1 = 7.

Step2: Find f(-2)

Substitute x=-2 into f(x)=8 - x², so f(-2)=8-(-2)²=8 - 4 = 4.

Step3: Solve part (A)

Substitute f(-1) and f(-2) into $\frac{f(-1)-f(-2)}{(-1)-(-2)}$, we get $\frac{7 - 4}{-1+2}=\frac{3}{1}=3$.

Step4: Find f(-2 + h)

Substitute x=-2 + h into f(x)=8 - x², f(-2 + h)=8-(-2 + h)²=8-(4 - 4h+h²)=8 - 4 + 4h - h²=4 + 4h - h².

Step5: Solve part (B)

Substitute f(-2 + h) and f(-2) into $\frac{f(-2 + h)-f(-2)}{h}$, we have $\frac{(4 + 4h - h²)-4}{h}=\frac{4h - h²}{h}=4 - h$.

Step6: Solve part (C)

We know from part (B) that $\frac{f(-2 + h)-f(-2)}{h}=4 - h$. Now find $\lim_{h
ightarrow0}\frac{f(-2 + h)-f(-2)}{h}$, substitute h = 0 into 4 - h, we get $\lim_{h
ightarrow0}(4 - h)=4$.

Answer:

(A) 3
(B) 4 - h
(C) 4