attempt 2: 1 attempt remaining. let f(x) = (5x^2 - 4)/(6x^2 + 8). evaluate f(x) at the following points: (a)…

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $f(x)=\frac{u(x)}{v(x)}$, then $f^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v(x)^2}$. Here, $u(x)=5x^{2}-4$, so $u^{\prime}(x) = 10x$, and $v(x)=6x^{2}+8$, so $v^{\prime}(x)=12x$. Then $f^{\prime}(x)=\frac{10x(6x^{2}+8)-(5x^{2}-4)\times12x}{(6x^{2}+8)^{2}}$.

Step2: Expand the numerator

Expand $10x(6x^{2}+8)-(5x^{2}-4)\times12x$: [ \begin{align*} &10x(6x^{2}+8)-(5x^{2}-4)\times12x\ =&60x^{3}+80x-(60x^{3}-48x)\ =&60x^{3}+80x - 60x^{3}+48x\ =&128x \end{align*} ] So $f^{\prime}(x)=\frac{128x}{(6x^{2}+8)^{2}}$.

Step3: Evaluate $f^{\prime}(10)$

Substitute $x = 10$ into $f^{\prime}(x)$: [ \begin{align*} f^{\prime}(10)&=\frac{128\times10}{(6\times10^{2}+8)^{2}}\ &=\frac{1280}{(600 + 8)^{2}}\ &=\frac{1280}{608^{2}}\ &=\frac{1280}{369664}\ &\approx0.00503 \end{align*} ]

Step4: Evaluate $f^{\prime}(-1)$

Substitute $x=-1$ into $f^{\prime}(x)$: [ \begin{align*} f^{\prime}(-1)&=\frac{128\times(-1)}{(6\times(-1)^{2}+8)^{2}}\ &=\frac{-128}{(6 + 8)^{2}}\ &=\frac{-128}{196}\ &\approx - 1.58025 \end{align*} ]

Answer:

(A) $f^{\prime}(10)\approx0.00503$ (B) $f^{\prime}(-1)\approx - 1.58025$