# Calculus I

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• E2P13
• #13246

4.2: Find the absolute minimum and maximum for the functions on the given intervals:

Find any critical points. Make a table with the critical points together with the boundary points and their respective values. The absolute extrema will be found among these values.

#16863

a. find deriv for x>2 y=(x-2)/x^2=x^(-1)-2x^(-2); y ‘ = -x^-2+4x^-3=(-x+4)/x^3 so critical point at x=4
find deriv for x<2 y=-(x-2)/x^2=-x^(-1)+2x^(-2); y ' = x^-2-4x^-3=(x-4)/x^3 but x=4 is not in consideration
So 4 values of x to check are boundaries 1 and 5, and crtical points 2(corner, no derivative) and 4.
x y
1 1
2 0
4 0.125
5 0.12
so min is 0 and max is 1.
b. sinx/x has deriv (xcosx-sinx)/x^2
xcosx-sinx=0 means x=tanx. Use calculator to look for intersection. First intersection after x=0 occurs around 4.5, way outside our interval, so we only need to look at the boundary points 0 and 1. At 0, y=1, at 1, y=sin(1)=0.841470985
so are min is sin(1) and our max is 1.

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