will the function f(x)=-1/(2x^3 - 4) be changing more rapidly at x = 2 or x = 6? why? * student can enter…

Explanation:

Step1: Find the derivative of $f(x)$

Use the quotient - rule. If $f(x)=\frac{u}{v}$ where $u = - 1$ and $v=2x^{3}-4$, then $u'=0$ and $v' = 6x^{2}$. The quotient - rule states that $f'(x)=\frac{u'v - uv'}{v^{2}}$. So $f'(x)=\frac{0\times(2x^{3}-4)-(-1)\times6x^{2}}{(2x^{3}-4)^{2}}=\frac{6x^{2}}{(2x^{3}-4)^{2}}$.

Step2: Evaluate $f'(x)$ at $x = 2$

Substitute $x = 2$ into $f'(x)$. $f'(2)=\frac{6\times2^{2}}{(2\times2^{3}-4)^{2}}=\frac{6\times4}{(16 - 4)^{2}}=\frac{24}{144}=\frac{1}{6}$.

Step3: Evaluate $f'(x)$ at $x = 6$

Substitute $x = 6$ into $f'(x)$. $f'(6)=\frac{6\times6^{2}}{(2\times6^{3}-4)^{2}}=\frac{6\times36}{(432 - 4)^{2}}=\frac{216}{428^{2}}=\frac{216}{183184}\approx0.00118$.

Answer:

The function $f(x)$ is changing more rapidly at $x = 2$ because $f'(2)=\frac{1}{6}\approx0.167$ and $f'(6)\approx0.00118$, and $\frac{1}{6}>0.00118$.