Answer to Question #208340 in Calculus for Sheen

Question

Find the absolute extrema of the function on the indicated

interval

1. 𝑔(𝑡) = 2 sec 1/2 𝑡 ; [ −1/3𝜋,1/2𝜋 ]

2. 𝑔(𝑥) = 𝑥³ + 3𝑥² − 9𝑥 ; [−4, 4]

3. 𝑓(𝑥) = 𝑥³ + 5𝑥 − 4 ; [ −3, −1 ]

4. ℎ(𝑥) = (𝑥 − 3)^1/3 + 4 ; [ 0, 2 ]

5. 𝑓(𝑡) = 3 cos 2𝑡 ; [ 1/6𝜋,3/4𝜋 ]

Answer 1

Solution.

1.

$g(t)=2\sec{\frac{1}{2}t}, [-\frac{1}{3}π,\frac{1}{2}π].$

We find the critical point of g. To do so, take the derivative:

$g'(t)=\sec{\frac{t}{2}}\tan{\frac{t}{2}}.$

Then $g'(t)=0$ when$\sec{\frac{t}{2}}\tan{\frac{t}{2}}=0,\newline \sec{\frac{t}{2}}=0\text{ or }\tan{\frac{t}{2}}=0,\newline t=2πk, k\in Z.$

$t=0 \in [-\frac{1}{3}π,\frac{1}{2}π].$

Evaluate g at the critical points and on the endpoints of the interval:

$g(0)=2,\newline g(-\frac{1}{3}π)=2\frac{1}{\cos{\frac{π}{6}}}=\frac{4}{\sqrt{3}},\newline g(\frac{1}{2}π)=2\frac{1}{\cos{\frac{π}{4}}}=\frac{4}{\sqrt{2}}.$

Therefore, g achieves its absolute minimum of 2 at t=0 and its absolute maximum of $\frac{4}{\sqrt{2}}$ at $t=\frac{1}{2}π.$ .

2.

$g(x)=x^3+3x^2-9x, [-4,4].$

We find the critical point of g. To do so, take the derivative:

$g'(x)=3x^2+6x-9.$

Then $g'(x)=0$ when$3x^2+6x-9=0,\newline x^2+2x-3=0,\newline x=-3, \text{or } x=1.$

Evaluate g at the critical points and on the endpoints of the interval:

$g(-3)=-27+27+27=27,\newline g(1)=1+3-9=-5,\newline g(-4)=-64+48+36=20,\newline g(4)=64+48-36=76.$

Therefore, g achieves its absolute minimum of -5 at x=1 and its absolute maximum of 76 at x=4.

3.

$f(x)=x^3+5x-4, [-3,-1].$

We find the critical point of f. To do so, take the derivative:

$f'(x)=3x^2+5.$

Then $f'(x)=0$ when$3x^2+5=0,\newline x^2=-\frac{5}{3},\newline x\in \varnothing.$

Evaluate f only on the endpoints of the interval:

$f(-3)=-27+15-4=-16,\newline f(-1)=-1+5-4=0.\newline$

Therefore, f achieves its absolute minimum of -16 at x=-3 and its absolute maximum of 0 at x=-1.

4.

$h(x)=(x-3)^{\frac{1}{3}}+4, [0,2].$

We find the critical point of h. To do so, take the derivative:

$h'(x)=\frac{1}{3}(x-3)^{-\frac{2}{3}}.$

Then $h'(x)=0$ when$\frac{1}{3}(x-3)^{-\frac{2}{3}}=0,\newline x=3.$

Evaluate h at the critical points and on the endpoints of the interval:

$h(3)=4,\newline h(0)=-\sqrt[3]{3}+4,\newline h(2)=-\sqrt[3]{1}+4=3.\newline$

Therefore, h achieves its absolute minimum of $-\sqrt[3]{3}+4$ at x=0 and its absolute maximum of 4 at x=3.

5.

$f(t)=3\cos{2t}, [\frac{1}{6}π,\frac{3}{4}π].$

We find the critical point of f. To do so, take the derivative:

$f'(x)=-6\sin{2t}.$

Then $f'(x)=0$ when$-6\sin{2t}=0,\newline \sin{2t}=0,\newline t=\frac{πk}{2}, k\in Z.$

$t=\frac{π}{2}\in [\frac{1}{6}π,\frac{3}{4}π].$

Evaluate f at the critical points and on the endpoints of the interval:

$f(\frac{π}{6})=3\cos{\frac{π}{3}}=1.5,\newline f(\frac{3π}{4})=3\cos{\frac{3π}{2}}=0,\newline f(\frac{π}{2})=3\cos{π}=-3.\newline$

Therefore, f achieves its absolute minimum of -3 at $t=\frac{π}{2},$ and its absolute maximum of 1.5 at $t=\frac{π}{6}.$