find the following using the table below.\n| $x$ | 1 | 2 | 3 | 4 |\n| $f(x)$ | 1 | | 4 | 3 |\n| $f(x)$ | 3 |…

find the following using the table below.\n| $x$ | 1 | 2 | 3 | 4 |\n| $f(x)$ | 1 | | 4 | 3 |\n| $f(x)$ | 3 | 1 | 2 | 4 |\n| $g(x)$ | 3 | 2 | 4 | 1 |\n| $g(x)$ | 2 | 4 | 1 | 3 |\n$h(1)$ if $h(x)=f(x)cdot g(x)$\n$h(1)$ if $h(x)=\frac{f(x)}{g(x)}$\n$h(1)$ if $h(x)=f(g(x))$\nquestion help: video\nsubmit question

find the following using the table below.\n| $x$ | 1 | 2 | 3 | 4 |\n| $f(x)$ | 1 | | 4 | 3 |\n| $f(x)$ | 3 | 1 | 2 | 4 |\n| $g(x)$ | 3 | 2 | 4 | 1 |\n| $g(x)$ | 2 | 4 | 1 | 3 |\n$h(1)$ if $h(x)=f(x)cdot g(x)$\n$h(1)$ if $h(x)=\frac{f(x)}{g(x)}$\n$h(1)$ if $h(x)=f(g(x))$\nquestion help: video\nsubmit question

Answer

Explanation:

Step1: Recall product - rule

If $h(x)=f(x)\cdot g(x)$, then $h^{\prime}(x)=f^{\prime}(x)g(x)+f(x)g^{\prime}(x)$. When $x = 1$, $h^{\prime}(1)=f^{\prime}(1)g(1)+f(1)g^{\prime}(1)$. From the table, $f^{\prime}(1) = 3$, $g(1)=3$, $f(1)=1$, $g^{\prime}(1)=2$. $h^{\prime}(1)=3\times3 + 1\times2=9 + 2=11$.

Step2: Recall quotient - rule

If $h(x)=\frac{f(x)}{g(x)}$, then $h^{\prime}(x)=\frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{g^{2}(x)}$. When $x = 1$, $h^{\prime}(1)=\frac{f^{\prime}(1)g(1)-f(1)g^{\prime}(1)}{g^{2}(1)}$. Substitute $f^{\prime}(1) = 3$, $g(1)=3$, $f(1)=1$, $g^{\prime}(1)=2$ into the formula. $h^{\prime}(1)=\frac{3\times3-1\times2}{3^{2}}=\frac{9 - 2}{9}=\frac{7}{9}$.

Step3: Recall chain - rule

If $h(x)=f(g(x))$, then $h^{\prime}(x)=f^{\prime}(g(x))\cdot g^{\prime}(x)$. When $x = 1$, first find $g(1)=3$. Then $h^{\prime}(1)=f^{\prime}(g(1))\cdot g^{\prime}(1)=f^{\prime}(3)\cdot g^{\prime}(1)$. From the table, $f^{\prime}(3)=2$ and $g^{\prime}(1)=2$. $h^{\prime}(1)=2\times2 = 4$.

Answer:

$h^{\prime}(1)$ when $h(x)=f(x)\cdot g(x)$ is $11$. $h^{\prime}(1)$ when $h(x)=\frac{f(x)}{g(x)}$ is $\frac{7}{9}$. $h^{\prime}(1)$ when $h(x)=f(g(x))$ is $4$.