Question Statement

Find the absolute extrema of the function on the indicated interval:

(i) $f(x)=-x^2+6x$ on $[1,4]$

(ii) $f(x)=(x-1{)}^2$ on $[2,5]$

(iii) $f(x)=x^{\frac{2}{3}}$ on $[-1,8]$

(iv) $f(x)=x^3-6x^2+2$ on $[-3,2]$

(v) $f(x)=1+5\sin 3x$ on $\left[0,\frac{\pi }{2}\right]$

(vi) $f(x)=2\cos 2x-4\cos x$ on $[0,2\pi ]$

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Background and Explanation

To find absolute extrema on a closed interval $[a,b]$, we apply the Closed Interval Method: identify all critical points (where $f^{\prime }(x)=0$ or $f^{\prime }(x)$ is undefined) that lie within the interval, then evaluate the function at these critical points and at both endpoints. The largest resulting value is the absolute maximum, and the smallest is the absolute minimum.

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Solution

Part (i): $f(x)=-x^2+6x$ on $[1,4]$

Given $f(x)=-x^2+6x$, we differentiate with respect to $x$:

$f^{\prime }(x)=-2x+6$

For critical values, we set $f^{\prime }(x)=0$:

$\begin{array}{rl}-2x+6& =0\\ -2x& =-6\\ x& =3\end{array}$

Since $x=3$ lies within the interval $[1,4]$, we evaluate the function at this critical point and at both endpoints:

$\begin{array}{rl}f(1)& =-(1{)}^2+6(1)=-1+6=5\\ f(3)& =-(3{)}^2+6(3)=-9+18=9\\ f(4)& =-(4{)}^2+6(4)=-16+24=8\end{array}$

Comparing these values: $5<8<9$.

Absolute minimum $=5$ at $x=1$

Absolute maximum $=9$ at $x=3$

Part (ii): $f(x)=(x-1{)}^2$ on $[2,5]$

Given $f(x)=(x-1{)}^2$, we differentiate using the chain rule:

$\begin{array}{rl}{f}^{\prime }(x)& =2(x-1{)}^{2-1}\cdot \frac{d}{dx}(x-1)\\ & =2(x-1)\cdot (1-0)\\ f^{\prime }(x)& =2(x-1)\end{array}$

For critical values, set $f^{\prime }(x)=0$:

$\begin{array}{rl}2(x-1)& =0\\ x-1& =0\\ x& =1\end{array}$

Since $x=1$ is not in the interval $[2,5]$, we only evaluate at the endpoints:

$\begin{array}{rl}f(2)& =(2-1{)}^2=(1{)}^2=1\\ f(5)& =(5-1{)}^2=(4{)}^2=16\end{array}$

Absolute Minimum $=1$ at $x=2$

Absolute Maximum $=16$ at $x=5$

Part (iii): $f(x)=x^{\frac{2}{3}}$ on $[-1,8]$

Given $f(x)=x^{\frac{2}{3}}$, we differentiate using the power rule:

$f^{\prime }(x)=\frac{2}{3}{x}^{\frac{2}{3}-1}=\frac{2}{3}{x}^{\frac{-1}{3}}=\frac{2}{3x^{1/3}}$

The derivative $f^{\prime }(x)$ is undefined at $x=0$, which lies within $[-1,8]$. Therefore, $x=0$ is a critical point.

Evaluating at the critical point and endpoints:

$\begin{array}{rl}f(-1)& =(-1{)}^{2/3}={\left((-1{)}^2\right)}^{1/3}=(1{)}^{1/3}=1\\ f(0)& =(0{)}^{2/3}=0\\ f(8)& =(8{)}^{2/3}={\left(2^3\right)}^{2/3}=2^2=4\end{array}$

Absolute Minimum $=0$ at $x=0$

Absolute Maximum $=4$ at $x=8$

Part (iv): $f(x)=x^3-6x^2+2$ on $[-3,2]$

Given $f(x)=x^3-6x^2+2$, we differentiate:

$\begin{array}{rl}{f}^{\prime }(x)& =3x^2-12x+0\\ f^{\prime }(x)& =3x^2-12x\end{array}$

For critical values, set $f^{\prime }(x)=0$:

$\begin{array}{ll}3x^2-12x=0& \\ 3x(x-4)=0& \\ 3x=0& x-4=0\\ x=0& x=4\notin [-3,2]\end{array}$

Since $x=4$ is not in the interval $[-3,2]$, we only consider $x=0$.

Evaluating at the critical point and endpoints:

$\begin{array}{rl}f(-3)& =(-3{)}^3-6(-3{)}^2+2=-27-54+2=-79\\ f(0)& =(0{)}^3-6(0{)}^2+2=0-0+2=2\\ f(2)& =(2{)}^3-6(2{)}^2+2=8-24+2=-14\end{array}$

Absolute Minimum $=-79$ at $x=-3$

Absolute Maximum $=2$ at $x=0$

Part (v): $f(x)=1+5\sin 3x$ on $\left[0,\frac{\pi }{2}\right]$

Given $f(x)=1+5\sin 3x$, we differentiate using the chain rule:

$\begin{array}{rl}{f}^{\prime }(x)& =0+5\cos 3x\cdot \frac{d}{dx}(3x)\\ & =5\cos 3x\cdot 3\\ f^{\prime }(x)& =15\cos 3x\end{array}$

For critical values, set $f^{\prime }(x)=0$:

$\begin{array}{rl}15\cos 3x& =0\\ \cos 3x& =0\\ 3x& =\frac{\pi }{2},\frac{3\pi }{2}\end{array}$

Solving for $x$ within $\left[0,\frac{\pi }{2}\right]$:

$\begin{array}{rr}3x=\frac{\pi }{2}& 3x=\frac{3\pi }{2}\\ x=\frac{\pi }{6}& x=\frac{\pi }{2}\end{array}$

Both values are in the interval. Evaluating at the critical points and endpoints:

$\begin{array}{rl}f(0)& =1+5\sin 0=1+0=1\\ f\left(\frac{\pi }{6}\right)& =1+5\sin \left(\frac{\pi }{2}\right)=1+5(1)=6\\ f\left(\frac{\pi }{2}\right)& =1+5\sin \left(\frac{3\pi }{2}\right)=1+5(-1)=-4\end{array}$

Absolute Minimum $=-4$ at $x=\frac{\pi }{2}$

Absolute Maximum $=6$ at $x=\frac{\pi }{6}$

Part (vi): $f(x)=2\cos 2x-4\cos x$ on $[0,2\pi ]$

Given $f(x)=2\cos 2x-4\cos x$, we differentiate:

$\begin{array}{rl}{f}^{\prime }(x)& =2\frac{d}{dx}(\cos 2x)-4\frac{d}{dx}(\cos x)\\ & =2(-\sin 2x)\cdot 2-4(-\sin x)\\ & =-4\sin 2x+4\sin x\\ & =-4(2\sin x\cos x)+4\sin x\\ & =-8\sin x\cos x+4\sin x\\ f^{\prime }(x)& =4\sin x(-2\cos x+1)=4\sin x(1-2\cos x)\end{array}$

For critical values, set $f^{\prime }(x)=0$:

$4\sin x(1-2\cos x)=0$

This yields two cases:

• $4\sin x=0\Rightarrow \sin x=0\Rightarrow x=0,\pi ,2\pi$
• $-2\cos x+1=0\Rightarrow \cos x=\frac{1}{2}\Rightarrow x=\frac{\pi }{3},\frac{5\pi }{3}$

Evaluating at all critical points and endpoints:

$\begin{array}{rl}f(0)& =2\cos 0-4\cos 0=2(1)-4(1)=-2\\ f\left(\frac{\pi }{3}\right)& =2\cos \left(\frac{2\pi }{3}\right)-4\cos \left(\frac{\pi }{3}\right)=2\left(-\frac{1}{2}\right)-4\left(\frac{1}{2}\right)=-1-2=-3\\ f(\pi )& =2\cos 2\pi -4\cos \pi =2(1)-4(-1)=2+4=6\\ f\left(\frac{5\pi }{3}\right)& =2\cos \left(\frac{10\pi }{3}\right)-4\cos \left(\frac{5\pi }{3}\right)=2\cos \left(\frac{4\pi }{3}\right)-4\left(\frac{1}{2}\right)=2\left(-\frac{1}{2}\right)-2=-3\\ f(2\pi )& =2\cos 4\pi -4\cos 2\pi =2(1)-4(1)=-2\end{array}$

Absolute minimum $=-3$ at $x=\frac{\pi }{3}$ and $x=\frac{5\pi }{3}$

Absolute maximum $=6$ at $x=\pi$

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Key Formulas or Methods Used

• Closed Interval Method: Evaluating function at critical points and endpoints to determine absolute extrema
• Power Rule: $\frac{d}{dx}[x^n]=n{x}^{n-1}$ for differentiation
• Chain Rule: $\frac{d}{dx}[f(g(x))]=f^{\prime }(g(x))\cdot g^{\prime }(x)$
• Trigonometric Derivatives: $\frac{d}{dx}[\sin (ax)]=a\cos (ax)$ and $\frac{d}{dx}[\cos (ax)]=-a\sin (ax)$
• Double Angle Identity: $\sin (2x)=2\sin (x)\cos (x)$ (used to simplify the derivative in part vi)
• Critical Points: Values in the domain where $f^{\prime }(x)=0$ or $f^{\prime }(x)$ is undefined

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Summary of Steps

1. Differentiate the function $f(x)$ to obtain $f^{\prime }(x)$
2. Find critical points by solving $f^{\prime }(x)=0$ and identifying where $f^{\prime }(x)$ is undefined
3. Filter critical points to keep only those within the given closed interval $[a,b]$
4. Evaluate the function $f(x)$ at all valid critical points and at both endpoints $x=a$ and $x=b$
5. Compare all calculated values: the largest is the absolute maximum, the smallest is the absolute minimum
6. State the final answer clearly, specifying both the extreme value and the specific $x$-value(s) where it occurs