let (f) and (g) be differentiable functions with the following properties: i. (f(x)<0) for all (x) ii. (g(5)…

let (f) and (g) be differentiable functions with the following properties: i. (f(x)<0) for all (x) ii. (g(5) = 2) if (h(x)=\frac{f(x)}{g(x)}) and (h(x)=\frac{f(x)}{g(x)}), then (g(x)=) (a) (\frac{1}{f(x)}) (b) (f(x)) (c) (-f(x)) (d) (0) (e) (2) the function (f) is defined by (f(x)=\frac{x}{x + 4}). what points ((x,y)) on the graph of (f) have the property that the line tangent to (f) at ((x,y)) has a slope of (\frac{1}{9})?

let (f) and (g) be differentiable functions with the following properties: i. (f(x)<0) for all (x) ii. (g(5) = 2) if (h(x)=\frac{f(x)}{g(x)}) and (h(x)=\frac{f(x)}{g(x)}), then (g(x)=) (a) (\frac{1}{f(x)}) (b) (f(x)) (c) (-f(x)) (d) (0) (e) (2) the function (f) is defined by (f(x)=\frac{x}{x + 4}). what points ((x,y)) on the graph of (f) have the property that the line tangent to (f) at ((x,y)) has a slope of (\frac{1}{9})?

Answer

Explanation:

Step1: Recall the quotient - rule

The quotient - rule states that if $h(x)=\frac{f(x)}{g(x)}$, then $h'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^{2}}$. Given that $h'(x)=\frac{f'(x)}{g(x)}$, we have $\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^{2}}=\frac{f'(x)}{g(x)}$.

Step2: Cross - multiply

Cross - multiplying the equation $\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^{2}}=\frac{f'(x)}{g(x)}$ gives $f'(x)g(x)-f(x)g'(x)=f'(x)g(x)$.

Step3: Simplify the equation

Subtracting $f'(x)g(x)$ from both sides of the equation $f'(x)g(x)-f(x)g'(x)=f'(x)g(x)$, we get $-f(x)g'(x)=0$. Since $f(x)<0$ for all $x$, then $g'(x) = 0$ for all $x$.

Step4: Find the function $g(x)$

If $g'(x)=0$ for all $x$, then $g(x)$ is a constant function. Given that $g(5)=2$, so $g(x)=2$ for all $x$.

Answer:

E. 2

Explanation for the second problem:

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

If $f(x)=\frac{x}{x + 4}$, using the quotient - rule $\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^{2}}$, where $u = x$, $u'=1$, $v=x + 4$, $v'=1$. Then $f'(x)=\frac{1\cdot(x + 4)-x\cdot1}{(x + 4)^{2}}=\frac{4}{(x + 4)^{2}}$.

Step2: Set the derivative equal to the given slope

We want to find $x$ such that $f'(x)=\frac{1}{9}$. So we set up the equation $\frac{4}{(x + 4)^{2}}=\frac{1}{9}$.

Step3: Cross - multiply and solve for $x$

Cross - multiplying gives $(x + 4)^{2}=36$. Taking the square root of both sides, we have $x + 4=\pm6$.

  • Case 1: If $x+4 = 6$, then $x = 2$.
  • Case 2: If $x + 4=-6$, then $x=-10$.

Step4: Find the corresponding $y$ values

When $x = 2$, $y=f(2)=\frac{2}{2 + 4}=\frac{1}{3}$. When $x=-10$, $y=f(-10)=\frac{-10}{-10 + 4}=\frac{5}{3}$.

Answer:

C. $(2,\frac{1}{3})$ and $(-10,\frac{5}{3})$