Question Statement

Use the second derivative test to find the relative extrema of the following functions:

(i) $f(x)=-(-2x-5{)}^2$

(ii) $f(x)=x^3+3x^2+3x+1$

(iii) $f(x)=6x^5-10x^2$

(iv) $f(x)=x^2+\frac{1}{x^2}$

(v) $f(x)=\cos 3x,[0,2\pi ]$

(vi) $f(x)=\cos x+\sin x,[0,2\pi ]$

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Background and Explanation

The second derivative test helps classify critical points: if $f^{\prime \prime }(c)>0$, the function has a relative minimum at $x=c$; if $f^{\prime \prime }(c)<0$, it has a relative maximum; if $f^{\prime \prime }(c)=0$, the test is inconclusive. First, find critical points by solving $f^{\prime }(x)=0$, then evaluate the second derivative at those points.

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Solution

Part (i): $f(x)=-(-2x-5{)}^2$

Given $f(x)=-(-2x-5{)}^2$.

Step 1: Find the first derivative using the chain rule. $\begin{array}{rl}{f}^{\prime }(x)& =-2(-2x-5{)}^{2-1}\cdot \frac{d}{dx}(-2x-5)\\ & =-2(-2x-5)\cdot (-2)\\ & =4(-2x-5)\\ & =-8x-20\end{array}$

Step 2: Find the second derivative. $\begin{array}{rl}{f}^{\prime \prime }(x)& =\frac{d}{dx}(-8x-20)\\ & =-8\end{array}$

Step 3: Find critical points by setting $f^{\prime }(x)=0$. $\begin{array}{rl}-8x-20& =0\\ -8x& =20\\ x& =-\frac{5}{2}\end{array}$

Step 4: Apply the second derivative test. At $x=-\frac{5}{2}$: $f^{\prime \prime }\left(-\frac{5}{2}\right)=-8<0$

Since $f^{\prime \prime }(x)<0$, the function has a relative maximum at $x=-\frac{5}{2}$.

Part (ii): $f(x)=x^3+3x^2+3x+1$

Given $f(x)=x^3+3x^2+3x+1$.

Step 1: Find the first derivative. $f^{\prime }(x)=3x^2+6x+3$

Step 2: Find the second derivative. $f^{\prime \prime }(x)=6x+6$

Step 3: Find critical points. $\begin{array}{rl}3x^2+6x+3& =0\\ 3(x+1{)}^2& =0\\ x& =-1\end{array}$

Step 4: Apply the second derivative test. At $x=-1$: $\begin{array}{rl}{f}^{\prime \prime }(-1)& =6(-1)+6\\ & =-6+6\\ & =0\end{array}$

Since $f^{\prime \prime }(-1)=0$, the second derivative test fails to determine the extrema.

Step 5: Find the point of inflection (since the test fails, we check the nature of the point). Substitute $x=-1$ into the original function: $\begin{array}{rl}f(-1)& =(-1{)}^3+3(-1{)}^2+3(-1)+1\\ & =-1+3-3+1\\ & =0\end{array}$

Therefore, the point of inflection is $(-1,0)$.

Part (iii): $f(x)=6x^5-10x^2$

Given $f(x)=6x^5-10x^2$.

Step 1: Find the first derivative. $f^{\prime }(x)=30x^4-20x$

Step 2: Find the second derivative. $f^{\prime \prime }(x)=120x^3-20$

Step 3: Find critical points. $\begin{array}{rl}30x^4-20x& =0\\ 10x(3x^3-2)& =0\end{array}$

This gives:

• $10x=0\Rightarrow x=0$
• $3x^3-2=0\Rightarrow x^3=\frac{2}{3}\Rightarrow x={\left(\frac{2}{3}\right)}^{\frac{1}{3}}$

Step 4: Apply the second derivative test at each critical point.

At $x=0$: $\begin{array}{rl}{f}^{\prime \prime }(0)& =120(0{)}^3-20\\ & =-20<0\end{array}$ Since $f^{\prime \prime }(0)<0$, $f(x)$ has a relative maximum at $x=0$.

At $x={\left(\frac{2}{3}\right)}^{\frac{1}{3}}$: $\begin{array}{rl}{f}^{\prime \prime }\left({\left(\frac{2}{3}\right)}^{\frac{1}{3}}\right)& =120{\left({\left(\frac{2}{3}\right)}^{\frac{1}{3}}\right)}^3-20\\ & =120\left(\frac{2}{3}\right)-20\\ & =80-20\\ & =60>0\end{array}$ Since $f^{\prime \prime }>0$, $f(x)$ has a relative minimum at $x={\left(\frac{2}{3}\right)}^{\frac{1}{3}}$.

Part (iv): $f(x)=x^2+\frac{1}{x^2}$

Given $f(x)=x^2+\frac{1}{x^2}=x^2+x^{-2}$.

Step 1: Find the first derivative. $\begin{array}{rl}{f}^{\prime }(x)& =2x+(-2)x^{-3}\\ & =2x-2x^{-3}\\ & =2x-\frac{2}{x^3}\end{array}$

Step 2: Find the second derivative. $\begin{array}{rl}{f}^{\prime \prime }(x)& =2+6x^{-4}\\ & =2+\frac{6}{x^4}\end{array}$

Step 3: Find critical points. $\begin{array}{rl}2x-\frac{2}{x^3}& =0\\ 2x& =\frac{2}{x^3}\\ x^4& =1\\ x^4-1& =0\\ (x^2-1)(x^2+1)& =0\end{array}$

This gives:

• $x^2-1=0\Rightarrow x=\pm 1$
• $x^2+1=0\Rightarrow x=\pm i$ (imaginary, so we discard these)

Step 4: Apply the second derivative test.

At $x=1$: $\begin{array}{rl}{f}^{\prime \prime }(1)& =2+\frac{6}{(1{)}^4}\\ & =2+6\\ & =8>0\end{array}$ Since $f^{\prime \prime }(1)>0$, $f(x)$ has a relative minimum at $x=1$.

At $x=-1$: $\begin{array}{rl}{f}^{\prime \prime }(-1)& =2+\frac{6}{(-1{)}^4}\\ & =2+6\\ & =8>0\end{array}$ Since $f^{\prime \prime }(-1)>0$, $f(x)$ has a relative minimum at $x=-1$.

Part (v): $f(x)=\cos 3x$ on $[0,2\pi ]$

Given $f(x)=\cos 3x$.

Step 1: Find the first derivative. $f^{\prime }(x)=-3\sin 3x$

Step 2: Find the second derivative. $\begin{array}{rl}{f}^{\prime \prime }(x)& =-3(3\cos 3x)\\ & =-9\cos 3x\end{array}$

Step 3: Find critical points in $[0,2\pi ]$. $\begin{array}{rl}-3\sin 3x& =0\\ \sin 3x& =0\\ 3x& =n\pi (n\in \mathbb{Z})\\ x& =\frac{n\pi }{3}\end{array}$

In the interval $[0,2\pi ]$, the critical points are: $x=0,\frac{\pi }{3},\frac{2\pi }{3},\pi ,\frac{4\pi }{3},\frac{5\pi }{3},2\pi$

Step 4: Evaluate $f^{\prime \prime }(x)$ at each critical point.

• At $x=0$: $f^{\prime \prime }(0)=-9\cos (0)=-9<0$ → Relative Maximum
• At $x=\frac{\pi }{3}$: $f^{\prime \prime }\left(\frac{\pi }{3}\right)=-9\cos (\pi )=-9(-1)=9>0$ → Relative Minimum
• At $x=\frac{2\pi }{3}$: $f^{\prime \prime }\left(\frac{2\pi }{3}\right)=-9\cos (2\pi )=-9(1)=-9<0$ → Relative Maximum
• At $x=\pi$: $f^{\prime \prime }(\pi )=-9\cos (3\pi )=-9(-1)=9>0$ → Relative Minimum
• At $x=\frac{4\pi }{3}$: $f^{\prime \prime }\left(\frac{4\pi }{3}\right)=-9\cos (4\pi )=-9(1)=-9<0$ → Relative Maximum
• At $x=\frac{5\pi }{3}$: $f^{\prime \prime }\left(\frac{5\pi }{3}\right)=-9\cos (5\pi )=-9(-1)=9>0$ → Relative Minimum
• At $x=2\pi$: $f^{\prime \prime }(2\pi )=-9\cos (6\pi )=-9(1)=-9<0$ → Relative Maximum

Part (vi): $f(x)=\cos x+\sin x$ on $[0,2\pi ]$

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

Step 1: Find the first derivative. $f^{\prime }(x)=-\sin x+\cos x$

Step 2: Find the second derivative. $\begin{array}{rl}{f}^{\prime \prime }(x)& =-\cos x-\sin x\end{array}$

Step 3: Find critical points. $\begin{array}{rl}-\sin x+\cos x& =0\\ \cos x& =\sin x\\ \tan x& =1\end{array}$

In $[0,2\pi ]$, the solutions are: $x=\frac{\pi }{4}\text{and}x=\frac{5\pi }{4}$

Step 4: Apply the second derivative test.

Case I: At $x=\frac{\pi }{4}$: $\begin{array}{rl}{f}^{\prime \prime }\left(\frac{\pi }{4}\right)& =-\cos \frac{\pi }{4}-\sin \frac{\pi }{4}\\ & =-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\\ & =-\frac{2}{\sqrt{2}}<0\end{array}$ Since $f^{\prime \prime }<0$, $f(x)$ has a relative maximum at $x=\frac{\pi }{4}$.

Case II: At $x=\frac{5\pi }{4}$: $\begin{array}{rl}{f}^{\prime \prime }\left(\frac{5\pi }{4}\right)& =-\cos \left(\frac{5\pi }{4}\right)-\sin \left(\frac{5\pi }{4}\right)\\ & =-\left(-\frac{1}{\sqrt{2}}\right)-\left(-\frac{1}{\sqrt{2}}\right)\\ & =\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\\ & =\frac{2}{\sqrt{2}}>0\end{array}$ Since $f^{\prime \prime }>0$, $f(x)$ has a relative minimum at $x=\frac{5\pi }{4}$.

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Key Formulas or Methods Used

• Chain Rule: $\frac{d}{dx}[f(g(x))]=f^{\prime }(g(x))\cdot g^{\prime }(x)$
• Power Rule: $\frac{d}{dx}(x^n)=n{x}^{n-1}$
• Trigonometric Derivatives: $\frac{d}{dx}(\sin x)=\cos x$, $\frac{d}{dx}(\cos x)=-\sin x$
• Second Derivative Test for Extrema:
  • If $f^{\prime }(c)=0$ and $f^{\prime \prime }(c)>0$: Relative minimum at $x=c$
  • If $f^{\prime }(c)=0$ and $f^{\prime \prime }(c)<0$: Relative maximum at $x=c$
  • If $f^{\prime }(c)=0$ and $f^{\prime \prime }(c)=0$: Test fails (inconclusive)
• Critical Points: Found by solving $f^{\prime }(x)=0$

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Summary of Steps

1. Differentiate the function to find $f^{\prime }(x)$ using appropriate rules (chain rule, power rule, etc.).
2. Differentiate again to find $f^{\prime \prime }(x)$.
3. Find critical points by setting $f^{\prime }(x)=0$ and solving for $x$.
4. Evaluate $f^{\prime \prime }(x)$ at each critical point $x=c$:
   • If $f^{\prime \prime }(c)>0$: Relative minimum
   • If $f^{\prime \prime }(c)<0$: Relative maximum
   • If $f^{\prime \prime }(c)=0$: Test fails; check for point of inflection or use first derivative test
5. For trigonometric functions on intervals: Find all solutions within the given domain before testing.