Finding the Absolute Minimum and Maximum It is a theorem that \"A continuous fun

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Finding the Absolute Minimum and Maximum

It is a theorem that "A continuous function f(x) on a closed interval [a, b] has an absolute maximum M and an absolute minimum m." This means there are some numbers c and d with m = f(c), M = f(d) and m <= f(x) <= M. Of course, the numbers c and d are limited to critical numbers or endpoints of the interval, a and b. As an example, we will find the absolute maximum and minimum of f( x ) = sin x + xcos( x 2 ) on the interval [ 0 , ? ]. syms x f(x) df(x) f(x) = sin(x)+x*cos(x^2); a=0;b=pi; hold off ezplot(f(x),[a,b]) hold on First, use the above graph and the cursor to find approximate values of the absolute maximum and minimum. Record them below to the nearest integer: M = m = Next, use the derivative to find exact location of the absolute maximum and minimum: Next, use the derivative to find exact location of the absolute maximum and minimum: df(x) =diff(f(x),x) ezplot(df(x), [a, b]); This function has a derivative at every point. Therefore, the only critical numbers are the solutions of the equation f '(x) = 0. solve(df(x),x); Matlab cannot find a general solution, so we will use the "fsolve" command to find decimal approximations to the solutions. From the graph, it is clear that there are four solutions of f ' (x) = 0, since the graph of f '(x) cuts the X-axis four times. By looking at the graph, verify that the four solutions to f' = 0 are in the four intervals [0.8, 1], [1.6, 2], [2.4, 2.6] and [3, PI] The commands below will find the four solutions, one in each interval. c1=vpasolve(df(x),x,[0.8 1]); c2=vpasolve(df(x),x,[1.6 2]); c3=vpasolve(df(x),x,[2.4 2.6]); c4=vpasolve(df(x),x,[3 pi]); Finally, we calculate the values of f at these four numbers and at 0 and ? (the endpoints of the interval under consideration): The apostrophe at the end of the command below lists the numbers in a column instead of a row. f(a) f(c1) f(c2) f(c3) f(c4) f(b)]' It's not too hard to identify the least (-3.01...) and greatest (3.100...) of the first 5 floating point numbers. Unfortuntately, it's not as easy to compare the value of f(b) to them. The vpa command will come in handy. vpa(f(b)) Now we can be sure the absolute maximum is f (c3) = 3.100075094, and the absolute minimum is f(c4) = -3.015500543.

Putting it All Together Now, repeat the above steps to find the absolute min and max of the function

(a) g(x) = x 4 ? 3 x 2 + x , x in [-1.5, 1.5],

and

(b) k(x) = x 3 4 ! ? sin(x) + 1 2 ! , x in [ 0 , 2 ? ]

Please post matlab code to find absolute min and max of the function using the above information

Explanation / Answer

f(x) = sin(x)+x*cos(x^2)

Deriving :
f'(x) = cosx + cos(x^2) + x*-sin(x^2)*(2x)

f'(x) = cosx + cos(x^2) - 2x^2*sin(x^2)

Now, for abs-max/min,
we do f' = 0

cosx + cos(x^2) - 2x^2*sin(x^2) = 0

Solving using graphing calculator,
we have
x = 0.92 , 1.825 , 2.51 and 3.09

Alongwith this,
we have endpoints x = 0 and x = pi

f(x) = sin(x)+x*cos(x^2)

When x= 0, f = 0
When x = pi, f = -2.8358697015517564
When x = 0.92, f = 1.40527
When x = 1.825, f = -0.82463
When x =2.51, f = 3.10007
When x = 3.09, f = -3.01496

So, abs max = 3.10007 when x = 2.51
And abs min = -3.01496 when x = 3.09

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