Find the minimum value of the function f(x)=2/x. + 3ln x on the interval [1/2, e]

Answer 1

Hi Danielle,

The extreme value theorem guarantees us that we must be able to find the absolute max and min of a continuous function on a closed interval. The first thing we need to do is find the critical points, to do this we need to take the derivative, set it equal to zero, and find what values of x it is true.

f(x) = 2/x + 3ln(x) → derivative → →f'(x) = -2/x² + 3/x = 0 → solve for x → x = 2/3

Now, if we think about what our function looks like, I think it is safe to assume that x = 2/3 is in fact a min, not a max; let's convince ourselves of this just to be sure. To do this we need to create 2 intervals, one on the left 2/3 and one on the right:

1/2 < x < 2/3 & 2/3 < x < e

We first need to pick a point in the left interval and plug it into the derivative, let's use x = 3/5:

f'(3/5) ≈ -.55

Since -.55 < 0, we know that the function is decreasing on this interval. Now we need to test the right interval, for this let's use x = 9/10:

f'(9/10) ≈ 0.86

Since 0.86 > 0, we know that the function is increasing on this interval.

We see that the interval on the left is decreasing and the interval on the right is increasing, this means that at x = 2/3 we must have the absolute minimum over the interval from [1/2, e]. Now, all that's left is to plug this value back into the original function.

f(2/3) ≈ 1.7836

We now wind up with our result that the minimum value of the function is 1.7836

Answer 2

To find the minimum value, one must take the derivative and set it = 0. The x's that you find are the critical numbers, it is at one of these x's (that cause the derivative to = 0) you will find the minimum or the maximum.

f(x) = 2/x + 3 ln(x)

f '(x) = -2/(x²) + 3/x getting a common denominator and setting this deriv = 0 we get

-2 + 3x = 0 Causing the numerator (top) to = 0 makes this derivative = 0, so -2 + 3x = 0 or x = 2/3

x²

Check that 1/2 <= x <= e and x = 2/3 falls within that interval. So the minimum is located at x = 2/3,

The minimum VALUE is the f(2/3) = the y-value when plugging x=2/3 back into the ORIGINAL equation.

f(2/3) = 2/(2/3) + 3 ln(2/3) = 3 + 3[ ln2 - ln3] ≈ 1.7836 is the minimum VALUE

(You can check this on a graphing calculator: graph f(x) = 2/x + 3 lnx, and ask your calculator to find the minimum: it finds it at 0.666666666666 with a value of 1.7836. So we know we got it right! :-)