a. Find the critical points of f on the given interval.
b. Determine the absolute extreme values off on the given interval
f(x)= x In(x/5); (0.1,5)
f = x ln(x/5)
f' = 1 + ln(x/5)
critical points where f'=0, so
1 + ln(x/5) = 0
x = 5/e
f(5/e) = -5/e
now, checking the endpoints of the interval
f(1/10) = -ln(50)/10
f(5) = 0
So, looks like the min is -5/e, max is 0.
To find the critical points of f on the given interval, we need to find the values of x where the derivative of f equals zero or is undefined.
a. To find the derivative of f, we can use the product rule:
f(x) = x * ln(x/5)
Using the product rule, the derivative of f(x) is given by:
f'(x) = 1 * ln(x/5) + x * (1 / (x/5)) * (1 / 5)
= ln(x/5) + 1/5
To find the critical points, we set the derivative equal to zero and solve for x:
ln(x/5) + 1/5 = 0
Subtracting 1/5 from both sides:
ln(x/5) = -1/5
Taking the exponential of both sides:
e^(ln(x/5)) = e^(-1/5)
x/5 = e^(-1/5)
Multiplying both sides by 5:
x = 5 * e^(-1/5)
So the critical point is x = 5 * e^(-1/5).
b. To determine the absolute extreme values of f on the given interval, we need to evaluate f(x) at the critical points and the endpoints of the interval (0.1 and 5), and then compare the values.
1. Evaluate f(x) at x = 0.1:
f(0.1) = 0.1 * ln(0.1/5)
2. Evaluate f(x) at x = 5 * e^(-1/5):
f(5 * e^(-1/5)) = (5 * e^(-1/5)) * ln((5 * e^(-1/5))/5)
3. Evaluate f(x) at x = 5:
f(5) = 5 * ln(5/5)
Now, compare the values of f(x) obtained from step 1, 2, and 3, and find the maximum and minimum values on the given interval.