Finding max/min on given interval....probably just algebra issues: f(t) = t - cbrt(t)

[Strat]

I've been going in circles on these 3 questions, they're all the same type. I've done a bunch of other ones correctly so I understand the concept and I assume it's the algebra I'm struggling with, but just in case I screwed something else up I've posted the whole problem. It's the part where I find where the derivative = 0 or DNE that I am struggling with.


Directions: Find the absolute maximum and absolute minimum values of f on the given interval:

1. f(t) = t - cbrt(t), [-1, 4]

My work: f'(t) = 1-(1/3 * t^-2/3)
(I know how to plug the endpoints -1 and 4 in to get there values so I am just trying to find where f'(t) = 0 or DNE)
0 = 1-((1/3 * t)^-2/3)
(1/3 * t)^-2/3= 1
(1/9)^1/3 * t^-2/3= 1
t^-2/3 = 1/((1/9)^1/3)
multiply by 1^-3/2
t = (1/((1/9)^1/3))^-3/2
t = 1/3

I've tried this a couple ways and always get 1/3 but t should = sqrt(3) / 9 and then the minimum value should be (-2 sqrt(3)) / 9.

2. f(x) = 2cosx + sin2x, [0, pi/2]

My work:
f'(x) = -2sinx +2cos2x (chain rule used)

0 = -sinx + cos2x
sinx = cos2x
I don't really know what I can do to this and the book is giving me pi/6 as an answer which corresponds to (3/2)(sqrt(3)). Did they just guess pi/6? I can't use a calculator and the only values I have memorized are 0, pi/2, pi, pi3/2.


3. f(x) = x^-2lnx, [1/2, 4]

My work:
f'(x) = (x^-2)(1/x) + (lnx)(-2x^-3)
f'(x) = (1/x³) + (lnx)/(-2x^-3)
f'(x) = (-2 + lnx)/(-2x³)

0 = (-2 + lnx)/(-2x³)
The obvious way to make this = 0 is make x = 0 but that is not in the interval so I don't know what else I can do. Somehow they get e^1/2​.

 

[MarkFL]

1.) \(\displaystyle f(t)=t-\sqrt[3]{t}\) on \(\displaystyle [-1,4]\)

\(\displaystyle \displaystyle f'(t)=1-\frac{1}{3}t^{-\frac{2}{3}}\)

\(\displaystyle \displaystyle f'(t)=\frac{3t^{\frac{2}{3}}-1}{3t^{\frac{2}{3}}}\)

Set the numerator to zero:

\(\displaystyle \displaystyle 3t^{\frac{2}{3}}-1=0\)

\(\displaystyle \displaystyle t^{\frac{2}{3}}=\frac{1}{3}\)

\(\displaystyle \displaystyle t^{\frac{1}{3}}=\pm\frac{1}{\sqrt{3}}\)

\(\displaystyle \displaystyle t=\pm\frac{1}{3\sqrt{3}}\)

Set the denominator to zero:

\(\displaystyle t=0\)

Now, you have 3 critical values and the 2 end-values of the given interval to check.

2.) \(\displaystyle f(x)=2\cos(x)+\sin(2x)\) on \(\displaystyle \displaystyle \left[0,\frac{\pi}{2}\right]\)

\(\displaystyle f'(x)=-2\sin(x)+2\cos(2x)=0\)

Divided through by 2, use a double-angle identity for cosine, and arrange in standard form:

\(\displaystyle 2\sin^2(x)+\sin(x)-1=0\)

Factor:

\(\displaystyle \left(2\sin(x)-1\right)\left(\sin(x)+1\right)=0\)

We know that on the given interval, the sine function is non-negative, thus we are left with:

\(\displaystyle \displaystyle \sin(x)=\frac{1}{2}\implies x=\frac{\pi}{6}\)

Counting the end-values, you now have 3 values to check.

 

[mmm4444bot]

3. f(x) = x^-2lnx, [1/2, 4]

My work:
f'(x) = (x^-2)(1/x) + (lnx)(-2x^-3)

f'(x) = (1/x³) + (lnx)/(-2x^-3) Division not correct; -2 goes on top

f'(x) = 1/x^3 - 2*ln(x)/x^3