Question

let ( f(x)=x^{3}). (a) estimate the values of ( f(0),f(\frac{1}{2}),f(1),f(2)), and ( f(3)) by zooming in on the graph of ( f). (round your answers to one - decimal place.) (b) use symmetry to deduce the values of ( f(-\frac{1}{2}),f(-1),f(-2)), and ( f(-3)). (round your answers to one - decimal place.) (c) use the values from parts (a) and (b) to graph ( f). (d) guess a formula for ( f(x)). use the definition of the derivative to verify that your guess is correct. ( f(x)=3x) ( f(x)=\frac{x^{4}}{4}) ( f(x)=3x^{2}) ( f(x)=3x + 2)

Explanation:

Step1: Recall derivative formula

The derivative of $f(x)=x^{3}$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ is $f'(x)=3x^{2}$.

Step2: Calculate $f'(x)$ at given positive points

For $x = 0$, $f'(0)=3\times0^{2}=0$.
For $x=\frac{1}{2}$, $f'(\frac{1}{2})=3\times(\frac{1}{2})^{2}=3\times\frac{1}{4}=0.75\approx0.8$.
For $x = 1$, $f'(1)=3\times1^{2}=3$.
For $x = 2$, $f'(2)=3\times2^{2}=3\times4 = 12$.
For $x = 3$, $f'(3)=3\times3^{2}=3\times9 = 27$.

Step3: Use symmetry of derivative

Since $f(x)=x^{3}$ is an odd function, its derivative $f'(x)=3x^{2}$ is an even function. So $f'(-x)=f'(x)$.
For $x=-\frac{1}{2}$, $f'(-\frac{1}{2})=f'(\frac{1}{2})\approx0.8$.
For $x=-1$, $f'(-1)=f'(1)=3$.
For $x=-2$, $f'(-2)=f'(2)=12$.
For $x=-3$, $f'(-3)=f'(3)=27$.

Step4: Graph $f'(x)$

The function $f'(x)=3x^{2}$ is a parabola opening upwards with vertex at the origin $(0,0)$.

Step5: Confirm derivative formula

The definition of the derivative is $f'(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$.
If $f(x)=x^{3}$, then $f(x + h)=(x + h)^{3}=x^{3}+3x^{2}h + 3xh^{2}+h^{3}$.
So $f'(x)=\lim_{h
ightarrow0}\frac{(x^{3}+3x^{2}h + 3xh^{2}+h^{3})-x^{3}}{h}=\lim_{h
ightarrow0}\frac{3x^{2}h+3xh^{2}+h^{3}}{h}=\lim_{h
ightarrow0}(3x^{2}+3xh + h^{2})=3x^{2}$.

Answer:

(a) $f'(0)=0$, $f'(\frac{1}{2})\approx0.8$, $f'(1)=3$, $f'(2)=12$, $f'(3)=27$
(b) $f'(-\frac{1}{2})\approx0.8$, $f'(-1)=3$, $f'(-2)=12$, $f'(-3)=27$
(c) The graph of $y = f'(x)=3x^{2}$ is a parabola opening upwards with vertex at $(0,0)$.
(d) The formula for $f'(x)$ is $f'(x)=3x^{2}$