question topic(s)/section(s): 2.9 the quotient rule\nclick the timer to pace yourself. some questions do not…

question topic(s)/section(s): 2.9 the quotient rule\nclick the timer to pace yourself. some questions do not require the full 2 minutes, and others will require a little more than 2 minutes.\n\n| $x$ | 0 | 3 | 5 | 9 |\n| ---- | ---- | ---- | ---- | ---- |\n| $f(x)$ | 3 | -6 | 9 | 2 |\n| $f(x)$ | 2 | 8 | 9 | 0 |\n| $g(x)$ | 5 | 4 | -1 | -4 |\n| $g(x)$ | 7 | -5 | -8 | 3 |\n\nthe functions $f$, $g$, $h$, and $k$ are twice - differentiable functions. let $h(x)=\frac{f(x)}{cos(x)}$. find $h(0)$.\n0\n3\n2\nundefined\nclear my selection

question topic(s)/section(s): 2.9 the quotient rule\nclick the timer to pace yourself. some questions do not require the full 2 minutes, and others will require a little more than 2 minutes.\n\n| $x$ | 0 | 3 | 5 | 9 |\n| ---- | ---- | ---- | ---- | ---- |\n| $f(x)$ | 3 | -6 | 9 | 2 |\n| $f(x)$ | 2 | 8 | 9 | 0 |\n| $g(x)$ | 5 | 4 | -1 | -4 |\n| $g(x)$ | 7 | -5 | -8 | 3 |\n\nthe functions $f$, $g$, $h$, and $k$ are twice - differentiable functions. let $h(x)=\frac{f(x)}{cos(x)}$. find $h(0)$.\n0\n3\n2\nundefined\nclear my selection

Answer

Explanation:

Step1: Recall the quotient - rule

The quotient - rule states that if $h(x)=\frac{u(x)}{v(x)}$, then $h^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v^{2}(x)}$. Here, $u(x) = f(x)$ and $v(x)=\cos(x)$. So $h^{\prime}(x)=\frac{f^{\prime}(x)\cos(x)-f(x)(-\sin(x))}{\cos^{2}(x)}=\frac{f^{\prime}(x)\cos(x)+f(x)\sin(x)}{\cos^{2}(x)}$.

Step2: Evaluate $h^{\prime}(0)$

Substitute $x = 0$ into the formula for $h^{\prime}(x)$. We know that $\cos(0)=1$, $\sin(0)=0$, $f(0) = 3$, and $f^{\prime}(0)=2$. $h^{\prime}(0)=\frac{f^{\prime}(0)\cos(0)+f(0)\sin(0)}{\cos^{2}(0)}$. Substitute the values: $h^{\prime}(0)=\frac{2\times1 + 3\times0}{1^{2}}$. $h^{\prime}(0)=\frac{2+0}{1}=2$.

Answer:

2