# Absolute maximum & minimum values.

• Jan 14th 2007, 05:50 PM
asnxbbyx113
Absolute 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 PM
CaptainBlack
Quote:

Originally Posted by asnxbbyx113
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.

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 PM
earboth
Quote:

Originally Posted by asnxbbyx113
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.

Hello,

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

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

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

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

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

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

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

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 PM
CaptainBlack
Quote:

Originally Posted by asnxbbyx113
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.

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