<p>How would you algebraically find the extreme values of the function </p>
<p>f(x) = 1/x + ln(x) in the interval 0.5 greater than equal to x greater than equal to 4?</p>
<p>the book's answer is:</p>
<p>Maximum value is 1/4 + ln4 at x=4
Minimum value is 1 at x=1
and local maximum at (1/2, 2-ln 2)</p>
<p>How on earth did they get this? I understand it graphically, but how do you do this, step-by-step please, algebraically? I'm desperate.</p>
<p>find the derivative of f(x) [which is -1/x^2 + 1/x]. then, you would find the critical points (where f'(x) does not equal 0 or dne)). it won't make a difference here since 0 isn't included in the restriction. then you make a 'sign chart.' f'(x) <------.5---------4-------> and you plug in numbers, so you'll 'see' the shape of the original graph. the original graph increases from .5 to 4, so the minimum value would be at x=.5 (plug this into the original equation) and the maximum at x=4 (also plug this into the original equation). not sure how they got the local maximum. </p>
<p>i believe i did this right....someone please correct me if i'm wrong.</p>
<p>(I assume you mean the interval 0.5 <= x <= 4, not 0.5 >= x >=4)</p>
<p>You can solve for f'(x) = 0. In this case,
(1/x) - (1/x^2) = 0
1/x = 1/x^2
x = x^2
or x=1</p>
<p>Compute f''(x) or d(f'(x))/dx as
f''(x) = (2/x^3) - (1/x^2)
At x=1, f''(1) = 2-1 = 1 > 0, so f(1) is a local minimum.</p>
<p>To find the maximum, evaluate f(x) at both ends of the interval and compare.
f(0.5) = 1/0.5 + ln(0.5) = 2 + ln(1/2) = 2 - ln(2)
f(4) = 1/4 + ln(2^2) = 0.25 + 2 ln(2)</p>
<p>Compare these two values, see which one is larger. Both f(0.5) and f(4) would be local maxima.</p>