Posted by George on Thursday, October 16, 2008 at 6:21pm.
I always encouraged my students to make a rough sketch of the graph.
Pick any 5 or 6 positive x's, then find their y values.
Pick an x close to zero (e.g. x = .00001) and see what y you get. This gives you an indication of the graph does close to zero. (remember x cannot be zero for this function)
In general, the max/min is found by setting the derivative equal to zero, solving for the variable, and then subbing that back into the original equation.
so
f(x) = 8x - ln(4x)
f'(x) = 8 - 1/x
set this equal to zero, you will get x = 1/8
I will let you find the minimum.
y = 8 x - ln (4x)
y' = 8 -4/x
when is that zero?
4/x = 8
x = 1/2
is that a maximum or a minimum?
y" = 0 +4/x^2
for x>0, y" is positive so it is a minimum
so minimum at x = 1/2
what is the value of the function at that minimum?
y = 8(1/2) - ln 2
4 - ln 2 = about 3.3
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