the graphs of the function f(x) and g(x) are given in blue and red, respectively. (click on a graph to…

Explanation:

Step1: Recall product - rule for $u(x)$

The product - rule states that if $u(x)=f(x)g(x)$, then $u^{\prime}(x)=f^{\prime}(x)g(x)+f(x)g^{\prime}(x)$. To find $u^{\prime}(1)$, we need to find $f(1)$, $f^{\prime}(1)$, $g(1)$ and $g^{\prime}(1)$ from the graphs.

Step2: Find function values at $x = 1$ from the graphs

From the graphs, when $x = 1$, $f(1)=2$ and $g(1)=2$.

Step3: Find slopes of tangent lines at $x = 1$ for $f(x)$ and $g(x)$

The slope of the tangent line to $y = f(x)$ at $x = 1$: For $f(x)$ (blue - graph), the slope $f^{\prime}(1)=2$. The slope of the tangent line to $y = g(x)$ at $x = 1$: For $g(x)$ (red - graph), the slope $g^{\prime}(1)=- 1$.

Step4: Calculate $u^{\prime}(1)$

Substitute into the product - rule formula: $u^{\prime}(1)=f^{\prime}(1)g(1)+f(1)g^{\prime}(1)$. So $u^{\prime}(1)=(2)\times(2)+(2)\times(-1)=4 - 2=2$.

Step5: Recall quotient - rule for $v(x)$

The quotient - rule states that if $v(x)=\frac{f(x)}{g(x)}$, then $v^{\prime}(x)=\frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{g^{2}(x)}$.

Step6: Calculate $v^{\prime}(1)$

Substitute $f(1) = 2$, $f^{\prime}(1)=2$, $g(1)=2$ and $g^{\prime}(1)=-1$ into the quotient - rule formula: $v^{\prime}(1)=\frac{(2)\times(2)-(2)\times(-1)}{2^{2}}=\frac{4 + 2}{4}=\frac{6}{4}=\frac{3}{2}$.

Answer:

$u^{\prime}(1)=2$ $v^{\prime}(1)=\frac{3}{2}$