Question

Find the rate of change for x3. You need to work out the change in f(x)=x3 when x is increased by a small number h to x+h. So you will work out f(x+h)-f(x). Then do some algebra to simplify this. Then divide this by h to get the average rate of change of f(x) between x and x+h.

Answer 1

Answer:

Explanation:

Given

$F(x)=x^3$

Rate of change of F(x) is given by

$F'(x)=\lim_{h\rightarrow 0}\\\frac{F(x+h)-F(x)}{x+h-x}$

$F'(x)=\lim_{h\rightarrow 0}\\\frac{(x+h)^3-x^3}{h}$

$F'(x)=\lim_{h\rightarrow 0}\\\frac{x^3+h^3+3x^2h+3xh^2-x^3}{h}$

$F'(x)=\lim_{h\rightarrow 0}\\\frac{h^3+3x^2h+3xh^2}{h}$

$F'(x)=\lim_{h\rightarrow 0}\\h^2+3x^2+3xh$

Putting limits

$F'(x)=3x^2$

                     

Answer 2

The average rate of change will be "3x²".

Average rate of change

According to the question,

The function, f(x) = x³

then,

f(x + h) = (x + h)³

Now,

→ f(x + h) - f(x) = (x + h)³ - x³

                      = x³ + h³ + 3x²h + 3xh² - x³

                      = h³ + 3x²h + 3xh²

and,

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

                    = $\frac{h[h^2+3x^2+3xh]}{h}$

                    = h² + 3x² + 3xh

By applying the limit, we get

→ $\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ = $\lim_{h \to 0}$ h² + 3x² + 3xh

By substituting the values,

                                = 0² + 3x² + 3x(0)

                                = 3x²

Thus the above approach is correct.      

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