# `f(x) = (4/3)xsqrt(3 - x), [0,3]` (a) Use a computer algebra system to graph the function and approximate any absolute extrema on the given interval. (b) Use the utility to find any critical...

`f(x) = (4/3)xsqrt(3 - x), [0,3]` (a) Use a computer algebra system to graph the function and approximate any absolute extrema on the given interval. (b) Use the utility to find any critical numbers, and use them to find any absolute extrema not located at the endpoints. Compare the results with those in part (a).

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See the attached graph plotted in Matlab.

a) approximate exterma

There is a maximum close to 2.

b)

>> f = (4/3)*x*sqrt(3-x)

f =

(4*x*(3 - x)^(1/2))/3

1. Asymptotes -

>> limit (f, inf)

ans =

Inf*1i

>> limit (f, -inf)

ans =

-Inf

This means there is no horizontal asymptote.

Vertical asymptote is at x=3

2. >> f1 = diff(f)

f1 =

(4*(3 - x)^(1/2))/3 - (2*x)/(3*(3 - x)^(1/2))

>> pretty(f1)

sqrt(3 - x) 4 2 x

------------- - -------------

3 3 sqrt(3 - x)

We can find the ciritical points by finding the roots of the derivative of f.

>> crit_pts = solve(f1)

crit_pts =

2

There is only one maximum [-inf, inf] and it is at x = 2, and y = 2.667.

The results in b matches with results in a.