Question

(a) find a simplified form of the difference quotient and (b) complete the following table
f(x) = 1 - 2x³

a) \\(\frac{f(x + h) - f(x)}{h} = \square\\)
(simplify your answer.)
b) fill in the table
\

$$\begin{tabular}{|c|c|c|} \\hline x & h & \\(\\frac{f(x + h) - f(x)}{h}\\) \\\\ \\hline 3 & 2 & \\\\ \\hline 3 & 1 & \\\\ \\hline 3 & 0.1 & \\\\ \\hline 3 & 0.01 & \\\\ \\hline \\end{tabular}$$

(type integers or decimals.)
also, there is a table on the right with x=3, h values 2,1,0.1,0.01 and the column \\(\frac{f(x + h) - f(x)}{h}\\)

Explanation:

Step1: Compute f(x+h)

$f(x+h) = 1 - 2(x+h)^3 = 1 - 2(x^3 + 3x^2h + 3xh^2 + h^3)$

Step2: Calculate f(x+h)-f(x)

$f(x+h)-f(x) = [1 - 2(x^3 + 3x^2h + 3xh^2 + h^3)] - (1 - 2x^3) = -6x^2h -6xh^2 -2h^3$

Step3: Simplify the difference quotient

$\frac{f(x+h)-f(x)}{h} = \frac{-6x^2h -6xh^2 -2h^3}{h} = -6x^2 -6xh -2h^2$

Step4: Substitute x=3 into the quotient

For x=3: $\frac{f(3+h)-f(3)}{h} = -6(3)^2 -6(3)h -2h^2 = -54 -18h -2h^2$

Step5: Compute values for h=2

$-54 -18(2) -2(2)^2 = -54 -36 -8 = -98$

Step6: Compute values for h=1

$-54 -18(1) -2(1)^2 = -54 -18 -2 = -74$

Step7: Compute values for h=0.1

$-54 -18(0.1) -2(0.1)^2 = -54 -1.8 -0.02 = -55.82$

Step8: Compute values for h=0.01

$-54 -18(0.01) -2(0.01)^2 = -54 -0.18 -0.0002 = -54.1802$

Answer:

a) $-6x^2 -6xh -2h^2$

b)

x	h	$\frac{f(x+h)-f(x)}{h}$
3	1	-74
3	0.1	-55.82
3	0.01	-54.1802