10. find the absolute max and min of $h(x)=2x^{3}+3x^{2}-72x + 3$ on $0,5$.\n11. find the absolute maximum…

Explanation:

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).

Answer:

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

Explanation:

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).

Answer:

The absolute maximum is (1) at (\theta=\frac{\pi}{2}) and the absolute minimum is (-1) at (\theta=-\frac{\pi}{2}).

Explanation:

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 (9) at (x = 1) and the absolute minimum is (5\times(-2)^{\frac{4}{5}}-8) at (x=-2).

Explanation:

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

For (f(x)=x^{2}-x + 2), using the power - rule ((x^n)'=nx^{n - 1}), (f'(x)=2x-1).

Step2: Find the critical point

Set (f'(x)=0), then (2x-1=0), so (x=\frac{1}{2}).

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

Evaluate (f(x)) at (x=\frac{1}{2}), (f(\frac{1}{2})=(\frac{1}{2})^{2}-\frac{1}{2}+2=\frac{1}{4}-\frac{1}{2}+2=\frac{1 - 2 + 8}{4}=\frac{7}{4}); at (x=\frac{1}{2}) (already considered), at (x = 2), (f(2)=2^{2}-2 + 2=4).

Answer:

The absolute maximum is (4) at (x = 2) and the absolute minimum is (\frac{7}{4}) at (x=\frac{1}{2}).