(1 point)\nresults for this submission\nentered answer preview result\n-3 -3 incorrect\n-1/4 -\\frac{1}{4}…

(1 point)\nresults for this submission\nentered answer preview result\n-3 -3 incorrect\n-1/4 -\\frac{1}{4} incorrect\nat least one of the answers above is not correct.\nnote: you can click on the graph to obtain a larger version in a new browser window.\nthe 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:\nu(1) =\nv(1) =\nnote: you can earn partial credit on this problem.\npreview my answers submit answers\nyour score was recorded.\nyou have attempted this problem 1 time.\nyou received a score of 0% for this attempt.\nyour overall recorded score is 0%.\nyou have unlimited attempts remaining.

(1 point)\nresults for this submission\nentered answer preview result\n-3 -3 incorrect\n-1/4 -\\frac{1}{4} incorrect\nat least one of the answers above is not correct.\nnote: you can click on the graph to obtain a larger version in a new browser window.\nthe 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:\nu(1) =\nv(1) =\nnote: you can earn partial credit on this problem.\npreview my answers submit answers\nyour score was recorded.\nyou have attempted this problem 1 time.\nyou received a score of 0% for this attempt.\nyour overall recorded score is 0%.\nyou have unlimited attempts remaining.

Answer

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:

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