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(θ)=sin(θ)-θ on -π/2,π/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 + 2/x on 1/2,2.

Explanation:

Step1: Find the derivative of the function

For \(h(x)=2x^{3}+3x^{2}-72x + 3\), using the power - rule \((x^n)^\prime=nx^{n - 1}\), we have \(h^\prime(x)=6x^{2}+6x - 72=6(x^{2}+x - 12)=6(x + 4)(x - 3)\).

Step2: Find the critical points

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

Step3: Evaluate the function at the critical point and endpoints

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

Step1: Find the derivative of the function

For \(g(\theta)=\sin(\theta)-\theta\), \(g^\prime(\theta)=\cos(\theta)-1\).

Step2: Find the critical points

Set \(g^\prime(\theta)=0\), so \(\cos(\theta)-1 = 0\), which gives \(\cos(\theta)=1\). In the interval \([-\frac{\pi}{2},\frac{\pi}{2}]\), \(\theta = 0\) is the solution.

Step3: Evaluate the function at the critical point and endpoints

Evaluate \(g(\theta)\) at \(\theta=-\frac{\pi}{2}\), \(g(-\frac{\pi}{2})=\sin(-\frac{\pi}{2})+\frac{\pi}{2}=-1+\frac{\pi}{2}\approx - 1+1.57 = 0.57\).
Evaluate \(g(\theta)\) at \(\theta = 0\), \(g(0)=\sin(0)-0=0\).
Evaluate \(g(\theta)\) at \(\theta=\frac{\pi}{2}\), \(g(\frac{\pi}{2})=\sin(\frac{\pi}{2})-\frac{\pi}{2}=1-\frac{\pi}{2}\approx1 - 1.57=-0.57\).

Step1: Find the derivative of the function

For \(f(x)=5x^{\frac{4}{5}}+4x\), \(f^\prime(x)=5\times\frac{4}{5}x^{-\frac{1}{5}}+4 = 4x^{-\frac{1}{5}}+4=\frac{4}{x^{\frac{1}{5}}}+4=\frac{4 + 4x^{\frac{1}{5}}}{x^{\frac{1}{5}}}\).

Step2: Find the critical points

Set \(f^\prime(x)=0\), \(\frac{4 + 4x^{\frac{1}{5}}}{x^{\frac{1}{5}}}=0\). The numerator \(4 + 4x^{\frac{1}{5}}=0\) gives \(x^{\frac{1}{5}}=-1\), so \(x=-1\). Also, \(f^\prime(x)\) is undefined at \(x = 0\). Both \(x=-1\) and \(x = 0\) are in the interval \([-2,1]\).

Step3: Evaluate the function at the critical points and endpoints

Evaluate \(f(x)\) at \(x=-2\), \(f(-2)=5(-2)^{\frac{4}{5}}+4(-2)=5\times2^{\frac{4}{5}}-8\approx5\times1.74-8 = 0.7\).
Evaluate \(f(x)\) at \(x=-1\), \(f(-1)=5(-1)^{\frac{4}{5}}+4(-1)=5 - 4 = 1\).
Evaluate \(f(x)\) at \(x = 0\), \(f(0)=5(0)^{\frac{4}{5}}+4(0)=0\).
Evaluate \(f(x)\) at \(x = 1\), \(f(1)=5(1)^{\frac{4}{5}}+4(1)=5 + 4 = 9\).

Step1: Find the derivative of the function

For \(f(x)=x^{2}+2x\), \(f^\prime(x)=2x + 2\).

Step2: Find the critical points

Set \(f^\prime(x)=0\), \(2x+2 = 0\), which gives \(x=-1\). But \(x=-1\) is not in the interval \([\frac{1}{2},2]\).

Step3: Evaluate the function at the endpoints

Evaluate \(f(x)\) at \(x=\frac{1}{2}\), \(f(\frac{1}{2})=(\frac{1}{2})^{2}+2\times\frac{1}{2}=\frac{1}{4}+1=\frac{5}{4}\).
Evaluate \(f(x)\) at \(x = 2\), \(f(2)=2^{2}+2\times2=4 + 4 = 8\).

Answer:

The absolute maximum of \(h(x)\) on \([0,5]\) is \(3\) at \(x = 0\) and the absolute minimum is \(-132\) at \(x = 3\).