Answer in Calculus for Sheen #208340

Find the absolute extrema of the function on the indicated

interval

1. ( ) = 2 sec 1/2 ; [ −1/3 ,1/2 ]

2. ( ) = ³ + 3 ² − 9 ; [−4, 4]

3. ( ) = ³ + 5 − 4 ; [ −3, −1 ]

4. ℎ( ) = ( − 3)^1/3 + 4 ; [ 0, 2 ]

5. ( ) = 3 cos 2 ; [ 1/6 ,3/4 ]

Solution.

1.

“g(t)=2sec{frac{1}{2}t}, [-frac{1}{3}u03c0,frac{1}{2}u03c0].”

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,newlinensec{frac{t}{2}}=0text{ or n }tan{frac{t}{2}}=0,newlinent=2u03c0k, kin Z.”

“t=0 in [-frac{1}{3}u03c0,frac{1}{2}u03c0].”

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

“g(0)=2,newlineng(-frac{1}{3}u03c0)=2frac{1}{cos{frac{u03c0}{6}}}=frac{4}{sqrt{3}},newlineng(frac{1}{2}u03c0)=2frac{1}{cos{frac{u03c0}{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}u03c0.” .

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,newlinenx^2+2x-3=0,newlinenx=-3, text{or } x=1.”

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

“g(-3)=-27+27+27=27,newlineng(1)=1+3-9=-5,newlineng(-4)=-64+48+36=20,newlineng(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,newlinenx^2=-frac{5}{3},newlinenxin varnothing.”

Evaluate f only on the endpoints of the interval:

“f(-3)=-27+15-4=-16,newlinenf(-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,newlinenx=3.”

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

“h(3)=4,newlinenh(0)=-sqrt[3]{3}+4,newlinenh(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)=3cos{2t}, [frac{1}{6}u03c0,frac{3}{4}u03c0].”

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

“f'(x)=-6sin{2t}.”

Then “f'(x)=0” when“-6sin{2t}=0,newlinensin{2t}=0,newlinent=frac{u03c0k}{2}, kin Z.”

“t=frac{u03c0}{2}in [frac{1}{6}u03c0,frac{3}{4}u03c0].”

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

“f(frac{u03c0}{6})=3cos{frac{u03c0}{3}}=1.5,newlinenf(frac{3u03c0}{4})=3cos{frac{3u03c0}{2}}=0,newlinenf(frac{u03c0}{2})=3cos{u03c0}=-3.newline”

Therefore, f achieves its absolute minimum of -3 at “t=frac{u03c0}{2},” and its absolute maximum of 1.5 at “t=frac{u03c0}{6}.”