Absolute Minimum and Maximum

Amybee · #1 $Old$ November 3rd, 2009, 12:33 PM

Find the Max and Min values of f on the interval [1/2, 2]

for the equation:

f(x) = x - lnx

I am stuck on this, I tried to find the derivative but I can't figure it out. Any help would be greatly appreciated. Thanks.

adkinsjr · #2 $Old$ November 3rd, 2009, 01:04 PM

Quote:

Originally Posted by Amybee $View Post$

Find the Max and Min values of f on the interval [1/2, 2]

for the equation:

f(x) = x - lnx

I am stuck on this, I tried to find the derivative but I can't figure it out. Any help would be greatly appreciated. Thanks.

The derivative of the function is just $f'(x)=1-\frac{1}{x}$ . To find the max and mins, set this equal to zero and solve.

$1-\frac{1}{x}=0$

These are the cricitcal points. The only point seems to be at $x=1$ . This is in the given interval, so just find $(1,f(1))$ . If look at the value of the derivative just after $x=1$ , for example $f'(2)=\frac{1}{2}$ , it's clear that the point $(1,f(1))$ is a minimum. Google the "first derivative test" if you're not familiar with this.

Edit: the point I made is actually incorrect if you read string6bean1977's reply. I should have used the "Closed Interval Method." However, the point $(1,f(1))$ is still a minimum.

string6bean1977 · #3 $Old$ November 3rd, 2009, 01:12 PM

I agree with adkinsjr, but would like to add that the values of x that satisfy f`(x)=0 must then be tested in f(x) to ensure continuity. The zeroes of f`(x) don't automatically make them "c" values. (critical) Additionally, undefined values, such as 0 in the denominator, may also qualify as critical numbers. Such as x=0, in this equation, makes it undefined. However, f(0) doesn't exist either (ln 0), so, in this case, 0 is NOT a "c" value.

Good luck!

BB

adkinsjr · #4 $Old$ November 3rd, 2009, 01:26 PM

Quote:

Originally Posted by string6bean1977 $View Post$

I agree with adkinsjr, but would like to add that the values of x that satisfy f`(x)=0 must then be tested in f(x) to ensure continuity. The zeroes of f`(x) don't automatically make them "c" values. (critical) Additionally, undefined values, such as 0 in the denominator, may also qualify as critical numbers. Such as x=0, in this equation, makes it undefined. However, f(0) doesn't exist either (ln 0), so, in this case, 0 is NOT a "c" value.

Good luck!

BB

oh, your right! I didn't notice that. It's a good thing someone mentioned that.