Question

1. find the absolute max and min of $h(x)=2x^{3}+3x^{2}-72x + 3$ on $0,5$.
2. find the absolute maximum and minimum of $g(\theta)=\sin(\theta)$ on $-\frac{\pi}{2},\frac{\pi}{2}$.
3. find the absolute maximum and minimum of $f(x)=5x^{4/5}+4x$ on $-2,1$.
4. find the absolute maximum and minimum of $f(x)=x^{2}-x + 2$ on $\frac{1}{2},2$

Explanation:

10.

Step1: Find the derivative of \(h(x)\)

First, find \(h'(x)\) for \(h(x)=2x^{3}+3x^{2}-72x + 3\). Using the power - rule \((x^n)'=nx^{n - 1}\), we have \(h'(x)=6x^{2}+6x - 72=6(x^{2}+x - 12)=6(x + 4)(x - 3)\).

Step2: Find the critical points

Set \(h'(x)=0\), so \(6(x + 4)(x - 3)=0\). The critical points are \(x=-4\) and \(x = 3\). But we are interested in the interval \([0,5]\). So we only consider \(x = 3\).

Step3: Evaluate \(h(x)\) at the critical point and endpoints

Evaluate \(h(x)\) at \(x = 0\), \(h(0)=3\); at \(x = 3\), \(h(3)=2\times3^{3}+3\times3^{2}-72\times3 + 3=2\times27+3\times9-216 + 3=54 + 27-216+3=-132\); at \(x = 5\), \(h(5)=2\times5^{3}+3\times5^{2}-72\times5 + 3=2\times125+3\times25-360 + 3=250+75-360 + 3=-32\).

Step1: Recall the range of the sine function

The function \(g(\theta)=\sin(\theta)\) has a derivative \(g'(\theta)=\cos(\theta)\). The critical points of \(y = \sin(\theta)\) are found by setting \(\cos(\theta)=0\). In the interval \([-\frac{\pi}{2},\frac{\pi}{2}]\), the critical point is \(\theta = 0\).

Step2: Evaluate \(g(\theta)\) at the critical point and endpoints

Evaluate \(g(\theta)\) at \(\theta=-\frac{\pi}{2}\), \(g(-\frac{\pi}{2})=-1\); at \(\theta = 0\), \(g(0)=0\); at \(\theta=\frac{\pi}{2}\), \(g(\frac{\pi}{2})=1\).

Step1: Find the derivative of \(f(x)\)

For \(f(x)=5x^{4/5}+4x\), using the power - rule \((x^n)'=nx^{n - 1}\), \(f'(x)=5\times\frac{4}{5}x^{-\frac{1}{5}}+4 = 4x^{-\frac{1}{5}}+4=\frac{4}{x^{\frac{1}{5}}}+4=\frac{4 + 4x^{\frac{1}{5}}}{x^{\frac{1}{5}}}\).

Step2: Find the critical points

Set \(f'(x)=0\), then \(\frac{4 + 4x^{\frac{1}{5}}}{x^{\frac{1}{5}}}=0\). This implies \(4+4x^{\frac{1}{5}}=0\), so \(x^{\frac{1}{5}}=-1\) and \(x=-1\). But our interval is \([-2,1]\). Also, \(f'(x)\) is undefined at \(x = 0\).

Step3: Evaluate \(f(x)\) at the critical points and endpoints

Evaluate \(f(x)\) at \(x=-2\), \(f(-2)=5\times(-2)^{\frac{4}{5}}+4\times(-2)\); at \(x = 0\), \(f(0)=0\); at \(x = 1\), \(f(1)=5\times1^{\frac{4}{5}}+4\times1=9\).

Answer:

The absolute maximum is \(3\) at \(x = 0\) and the absolute minimum is \(-132\) at \(x = 3\).

11.