hw11 differentiation rules ii (target c1, c2, c5; §3.3)\nscore: 6/8 answered: 6/8\nquestion 7\nif…

Explanation:

Step1: Find slope - form of $g(x)$

The line $g(x)$ passes through $(0,9)$ and $(5,4)$. Using the slope - formula $m=\frac{y_2 - y_1}{x_2 - x_1}$, we have $m=\frac{4 - 9}{5-0}=\frac{- 5}{5}=-1$. Using the point - slope form $y - y_1=m(x - x_1)$ with the point $(0,9)$, $g(x)=-x + 9$. Then $g'(x)=-1$.

Step2: Find slope - form of $h(x)$

For $h(x)$, in the interval $0\leq x\leq3$, it passes through $(0,0)$ and $(3,3)$. The slope $m_1=\frac{3 - 0}{3 - 0}=1$, so $h(x)=x$ for $0\leq x\leq3$ and $h'(x)=1$ for $0\leq x\leq3$.

Step3: Apply the product rule

The product rule states that if $f(x)=g(x)\cdot h(x)$, then $f'(x)=g'(x)h(x)+g(x)h'(x)$. We want to find $f'(1)$. When $x = 1$, $g(1)=-1 + 9=8$, $h(1)=1$, $g'(1)=-1$, $h'(1)=1$. Substitute these values into the product - rule formula: $f'(1)=g'(1)h(1)+g(1)h'(1)$. $f'(1)=(-1)\times1+8\times1$. $f'(1)=-1 + 8=7$.

Answer:

$7$