Consider the function f(x)=x−5ln(x), 1/5≤x≤10.
The absolute maximum value is =

and this occurs at x equals =

The absolute minimum value is =

and this occurs at x equals =

f(x) = x - 5 ln(x)

Get the derivative:

f'(x) = 1 - 5/x

Find the critical points:

• f'(x) = 1 - 5/x = 0   ->   5/x = 1   ->   x = 5

• f'(x) is undefined for x = 0, but that's outside the domain of f(x)

Get the second derivative:

f''(x) = 5/x ^2

Check the sign of the second derivative at the critical point x = 5:

f'' (5) = 5/5^2 = 5/25 = 1/5 > 0

Since the second derivative is positive at this point, this means x = 5 is a local minimum.

Check the value of f(x) at the critical point as well as the endpoints of the given domain:

f (5) = 5 - 5 ln(5) ≈ -3.407

f (1/5) = 1/5 - 5 ln(1/5) ≈ 1.809

f (10) = 10 - 5 ln(10) ≈ -1.513

So we have

• absolute maximum = 1/5 - 5 ln(1/5) at x = 1/5

• absolute minimum = 5 - 5 ln(5) at x = 5

The absolute maximum value is  f(5)=5−5ln(5), and this occurs at x equals 5

The absolute minimum value is  f(1/5)= 1/5−5ln(1/5), and this occurs at x equals 1/5

Given the function f(x)=x−5ln(x),

We are to find the absolute minimum and maximum value of the function

At the turning point f'(x) = x - 5ln(x)

f'(x) = 1 - 5/x

0 = 1 - 5/x

-5/x = -1

-x = -5

x = 5

Find the second derivative of the function;

f''(x) = -5/x²

Substitute the value of x = 5 into the second derivative function to have:

f''(5) = -5/5² = 1/5

Since f''(5) = 1/5 < 0, this is the local maximum and the point x = 5 is the absolute minimum

The absolute maximum value is  f(5)=5−5ln(5), and this occurs at x equals 5

The absolute minimum value is  f(1/5)= 1/5−5ln(1/5), and this occurs at x equals 1/5

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