Question Statement

Determine whether the given function has a relative extremum at the indicated points:

(i) $f(x)=\cos x\sin x,x=\frac{\pi }{4}$

(ii) $f(x)=x\sin x,x=0$

(iii) $f(x)={\tan }^{2}x,x=\pi$

(iv) $f(x)=(1+\sin x{)}^3,x=\frac{\pi }{8}$

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Background and Explanation

To classify relative extrema, we use the second derivative test: if $f^{\prime }(c)=0$ and $f^{\prime \prime }(c)<0$, then $f$ has a relative maximum at $c$; if $f^{\prime \prime }(c)>0$, it has a relative minimum. First, we must verify that the point is a critical point (where $f^{\prime }(x)=0$ or undefined).

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Solution

Part (i): $f(x)=\cos x\sin x$ at $x=\frac{\pi }{4}$

Given $f(x)=\cos x\sin x$.

Differentiate with respect to $x$ using the product rule: $\begin{array}{rl}{f}^{\prime }(x)& =\frac{d}{dx}(\cos x)\cdot \sin x+\cos x\cdot \frac{d}{dx}(\sin x)\\ & =-\sin x\cdot \sin x+\cos x\cdot \cos x\\ & =-{\sin }^{2}x+{\cos }^{2}x\\ & ={\cos }^{2}x-{\sin }^{2}x\\ & =\cos 2x\end{array}$

Differentiate again to find the second derivative: $\begin{array}{rl}{f}^{\prime \prime }(x)& =\frac{d}{dx}(\cos 2x)\\ & =-\sin (2x)\cdot 2\\ & =-2\sin 2x\end{array}$

Evaluate at $x=\frac{\pi }{4}$: $\begin{array}{rl}{f}^{\prime \prime }\left(\frac{\pi }{4}\right)& =-2\sin \left(2\cdot \frac{\pi }{4}\right)\\ & =-2\sin \left(\frac{\pi }{2}\right)\\ & =-2(1)\\ & =-2<0\end{array}$

Since $f^{\prime \prime }\left(\frac{\pi }{4}\right)=-2<0$, the function $f(x)$ has a relative maximum at $x=\frac{\pi }{4}$.

Part (ii): $f(x)=x\sin x$ at $x=0$

Given $f(x)=x\sin x$.

Differentiate with respect to $x$ using the product rule: $\begin{array}{rl}{f}^{\prime }(x)& =1\cdot \sin x+x\cdot \cos x\\ & =\sin x+x\cos x\end{array}$

Differentiate again (applying product rule to the second term): $\begin{array}{rl}{f}^{\prime \prime }(x)& =\cos x+(1\cdot \cos x+x(-\sin x))\\ & =\cos x+\cos x-x\sin x\\ & =2\cos x-x\sin x\end{array}$

Evaluate at $x=0$: $\begin{array}{rl}{f}^{\prime \prime }(0)& =2\cos (0)-0\cdot \sin (0)\\ & =2(1)-0\\ & =2>0\end{array}$

Since $f^{\prime \prime }(0)=2>0$, the function $f(x)$ has a relative minimum at $x=0$.

Part (iii): $f(x)={\tan }^{2}x$ at $x=\pi$

Given $f(x)={\tan }^{2}x$.

Differentiate with respect to $x$ using the chain rule: $f^{\prime }(x)=2\tan x\cdot {\sec }^{2}x$

At $x=\pi$: $\begin{array}{rl}{f}^{\prime }(\pi )& =2\tan (\pi )\cdot {\sec }^2(\pi )\\ & =2(0)(1)\\ & =0\end{array}$

Since $f^{\prime }(\pi )=0$, this is a critical point. We proceed to find the second derivative.

Rewriting $f^{\prime }(x)$ using the identity ${\sec }^{2}x=1+{\tan }^{2}x$: $\begin{array}{rl}{f}^{\prime }(x)& =2\tan x(1+{\tan }^{2}x)\\ & =2\tan x+2{\tan }^{3}x\end{array}$

Differentiate again: $\begin{array}{rl}{f}^{\prime \prime }(x)& =2{\sec }^{2}x+2(3{\tan }^{2}x\cdot {\sec }^{2}x)\\ & =2{\sec }^{2}x+6{\tan }^{2}x{\sec }^{2}x\\ & =2{\sec }^{2}x(1+3{\tan }^{2}x)\end{array}$

Evaluate at $x=\pi$: $\begin{array}{rl}{f}^{\prime \prime }(\pi )& =2{\sec }^2(\pi )(1+3{\tan }^2(\pi ))\\ & =2(-1{)}^2(1+3(0))\\ & =2(1)(1)\\ & =2>0\end{array}$

Since $f^{\prime \prime }(\pi )=2>0$, the function $f(x)$ has a relative minimum at $x=\pi$.

Part (iv): $f(x)=(1+\sin x{)}^3$ at $x=\frac{\pi }{8}$

Given $f(x)=(1+\sin x{)}^3$.

Differentiate with respect to $x$ using the chain rule: $\begin{array}{rl}{f}^{\prime }(x)& =3(1+\sin x{)}^2\cdot \frac{d}{dx}(1+\sin x)\\ & =3(1+\sin x{)}^2\cdot \cos x\end{array}$

At $x=\frac{\pi }{8}$: Since $\frac{\pi }{8}$ is in the first quadrant ($0<\frac{\pi }{8}<\frac{\pi }{2}$), we have $\sin \left(\frac{\pi }{8}\right)>0$ and $\cos \left(\frac{\pi }{8}\right)>0$.

Therefore: $f^{\prime }\left(\frac{\pi }{8}\right)=3(1+\sin (\frac{\pi }{8}){)}^2\cdot \cos (\frac{\pi }{8})>0$

Since $f^{\prime }\left(\frac{\pi }{8}\right)\ne 0$, the point $x=\frac{\pi }{8}$ is not a critical point.

Hence, $f(x)$ has no relative extremum at $x=\frac{\pi }{8}$.

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Key Formulas or Methods Used

• Product Rule: $\frac{d}{dx}[u(x)v(x)]=u^{\prime }(x)v(x)+u(x)v^{\prime }(x)$
• Chain Rule: $\frac{d}{dx}[f(g(x))]=f^{\prime }(g(x))\cdot g^{\prime }(x)$
• Second Derivative Test: If $f^{\prime }(c)=0$ and $f^{\prime \prime }(c)<0\Rightarrow$ relative maximum; if $f^{\prime \prime }(c)>0\Rightarrow$ relative minimum
• Trigonometric Identities: ${\cos }^{2}x-{\sin }^{2}x=\cos 2x$, ${\sec }^{2}x=1+{\tan }^{2}x$, $\tan (\pi )=0$, $\sec (\pi )=-1$

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Summary of Steps

1. Compute the first derivative $f^{\prime }(x)$ using appropriate differentiation rules (product, chain, or power rules).
2. Verify critical point: Check if $f^{\prime }(c)=0$ at the indicated point. If $f^{\prime }(c)\ne 0$, conclude there is no relative extremum.
3. Compute the second derivative $f^{\prime \prime }(x)$ by differentiating $f^{\prime }(x)$.
4. Apply the second derivative test: Evaluate $f^{\prime \prime }(c)$. If negative $\Rightarrow$ relative maximum; if positive $\Rightarrow$ relative minimum.
5. State the conclusion clearly for each function at the indicated point.