Find the absolute maximum & minimum values of f(t)=2t(16-t^2)^(1/2) in the interval [-2, 4]. Give the answer in the form of f(a1)=b1, f(a2)=b2. Do not round the answers.

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- Jan 14th 2007, 05:50 PMasnxbbyx113Absolute maximum & minimum values.
Find the absolute maximum & minimum values of f(t)=2t(16-t^2)^(1/2) in the interval [-2, 4]. Give the answer in the form of f(a1)=b1, f(a2)=b2. Do not round the answers.

- Jan 14th 2007, 09:37 PMCaptainBlack
Procedure, find all the staionary points of f(t) (points at f'(t)=0). These

will include all the local maxima, minima, and some of the points of inflection.

Evaluate f(t) at these points, then the global maxima, will be the largest of

these and the value of f(t) at the end points of [-2,4], similarly for the

global minima.

The derivative of f is:

f'(t)= 2 sqrt(16-t^2)-2t^2/sqrt(16-t^2)

and f'(t)=0 has roots at +/-2 sqrt(2), the negative root is outside the

interval of interest, and f(2\sqrt(2))=16. (note possible exceptional points

at t=+/-4 which may need to be handled seperatly)

Now at the end points of the inteval we have f(-2)=-8 sqrt(3), and f(4)=0.

Hence the global maximum on [-2,4] is at t=2sqrt(2), and is 16, and the global

minimum is at t=-2 and is -8sqrt(3).

The attachment shows a plot of the function over the specified interval

RonL - Jan 14th 2007, 09:38 PMearboth
Hello,

calculate the 1rst derivative first (use product rule and chain rule):

$\displaystyle f'(t)=\sqrt{16-t^2} \cdot 2+2t \cdot \frac{1}{2} \cdot (16-t^2)^{-\frac{1}{2}} \cdot (-2t)=\frac{(16-t^2) \cdot 2-2t^2}{\sqrt{16-t^2}}$

f'(t) = 0 only if the numerator equals zero:

$\displaystyle (16-t^2) \cdot 2-2t^2=0 \Longleftrightarrow t=-2\ \vee \ t=2$

Plug in the values into the given equation of the function:

$\displaystyle f(-2)=2 \cdot (-2) \cdot \sqrt{16-(-2)^2}=-4 \cdot \sqrt{12}=-8\sqrt{3}$ or

$\displaystyle f(2)=2 \cdot (2) \cdot \sqrt{16-(2)^2}=4 \cdot \sqrt{12}=8\sqrt{3}$

Because one value is negative and the other is positive you can decide easily which one is the maximum and which one is the minimum without calculating the 2nd drivative.

EB - Jan 14th 2007, 10:37 PMCaptainBlack
I should have given a better summary of the principle involved here. It is that

the global maximum of a function on a closed interval is either a calculus type

local maximum where f'(t)=0, or it occurs at one of the end points of the

interval.

Similarly the global minimum of a function on a closed interval is either a

calculus type local minimum where f'(t)=0, or it occurs at one of the end points of the interval.

RonL