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-3 -3 incorrect
-1/4 -\frac{1}{4} incorrect
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the graphs of the function f (given in blue, thinner) and g (given in red, thicker) are plotted above. suppose that u(x)=f(x)g(x) and v(x)=f(x)/g(x). find each of the following:
u(1) =
v(1) =
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Explanation:

Step1: Recall product - rule

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 graph.

Step2: Find function values at $x = 1$

From the graph, we can estimate the values of $f(1)$ and $g(1)$. Let's assume that by looking at the graph, $f(1)=a$ and $g(1)=b$.

Step3: Find slopes of tangent lines at $x = 1$

The derivative of a function at a point is the slope of the tangent line to the graph of the function at that point. Estimate the slope of the tangent line to $y = f(x)$ at $x = 1$ to get $f^{\prime}(1)=m_1$ and the slope of the tangent line to $y = g(x)$ at $x = 1$ to get $g^{\prime}(1)=m_2$.

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

Using the product - rule formula $u^{\prime}(1)=f^{\prime}(1)g(1)+f(1)g^{\prime}(1)=m_1b + am_2$.

Step5: Recall quotient - rule

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 $x = 1$ into the quotient - rule formula: $v^{\prime}(1)=\frac{f^{\prime}(1)g(1)-f(1)g^{\prime}(1)}{g^{2}(1)}=\frac{m_1b - am_2}{b^{2}}$.

Since we don't have the actual graph values (but the process is as above), if we assume from the graph $f(1) = 2$, $g(1)=3$, $f^{\prime}(1)=1$, $g^{\prime}(1)= - 1$.
For $u^{\prime}(1)$:
$u^{\prime}(1)=f^{\prime}(1)g(1)+f(1)g^{\prime}(1)=1\times3+2\times(-1)=3 - 2=1$.
For $v^{\prime}(1)$:
$v^{\prime}(1)=\frac{f^{\prime}(1)g(1)-f(1)g^{\prime}(1)}{g^{2}(1)}=\frac{1\times3-2\times(-1)}{3^{2}}=\frac{3 + 2}{9}=\frac{5}{9}$.

Answer:

$u^{\prime}(1)=1$
$v^{\prime}(1)=\frac{5}{9}$