Question

ww4: problem 18 (1 point) results for this submission entered answer preview result 13 13 correct -0.75 -0.75 incorrect at least one of the answers above is not correct. 1 of the questions remains unanswered. the following table gives the values for functions ( f ) and ( g ) and their derivatives for integer values of ( x ) between 1 and 5. (\begin{array}{|c|c|c|c|c|c|}hline x&1&2&3&4&5\hline f(x)& - 1&-4&4&2&1\hline f(x)&-5&-3&-5&4&1\hline g(x)&-1&1&-1&1&-4\hline g(x)&5&-4&-2&2&-5\hlineend{array}) let ( p(x)=f(x)g(x),quad q(x)=\frac{f(x)}{g(x)}quad r(x)=xf(x)+\frac{g(x)}{x} ) find a) ( p(2)= ) 13 b) ( q(3)= ) c) ( r(4)= ) note: you can earn partial credit on this problem.

Explanation:

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

The product - rule states that if $p(x)=f(x)g(x)$, then $p^{\prime}(x)=f^{\prime}(x)g(x)+f(x)g^{\prime}(x)$. For $x = 2$, we have $f(2)=-4$, $f^{\prime}(2)=-3$, $g(2)=1$, and $g^{\prime}(2)=-4$. Then $p^{\prime}(2)=f^{\prime}(2)g(2)+f(2)g^{\prime}(2)=(-3)\times1+(-4)\times(-4)=-3 + 16=13$.

Step2: Recall quotient - rule for $q(x)$

The quotient - rule states that if $q(x)=\frac{f(x)}{g(x)}$, then $q^{\prime}(x)=\frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{g^{2}(x)}$. For $x = 3$, we have $f(3)=4$, $f^{\prime}(3)=-5$, $g(3)=-1$, and $g^{\prime}(3)=-2$. Then $q^{\prime}(3)=\frac{(-5)\times(-1)-4\times(-2)}{(-1)^{2}}=\frac{5 + 8}{1}=13$.

Step3: Recall sum - rule and product - rule for $r(x)$

First, rewrite $r(x)=xf(x)+\frac{g(x)}{x}=xf(x)+x^{-1}g(x)$. Using the sum - rule $(u + v)^{\prime}=u^{\prime}+v^{\prime}$, product - rule $(uv)^{\prime}=u^{\prime}v+uv^{\prime}$ for $u = x,v = f(x)$ and $u = x^{-1},v = g(x)$. The derivative of $xf(x)$ is $f(x)+xf^{\prime}(x)$ and the derivative of $\frac{g(x)}{x}$ is $\frac{g^{\prime}(x)x - g(x)}{x^{2}}$. So $r^{\prime}(x)=f(x)+xf^{\prime}(x)+\frac{g^{\prime}(x)x - g(x)}{x^{2}}$. For $x = 4$, we have $f(4)=2$, $f^{\prime}(4)=4$, $g(4)=1$, and $g^{\prime}(4)=2$. Then $r^{\prime}(4)=2+4\times4+\frac{2\times4 - 1}{4^{2}}=2 + 16+\frac{8 - 1}{16}=18+\frac{7}{16}=\frac{288 + 7}{16}=\frac{295}{16}=18.4375$.

Answer:

a) $13$
b) $13$
c) $18.4375$