Derivative?

Question

Use the given derivative to find all critical numbers of f, and at each critical number determine whether a rel. max,min, or neither occurs. f'(x)= (e^x-2)(e^x+3)

Answer 1

First set f'(x) = 0: that means e^x = 2, so x = ln 2.
You could differentiate again to check what type it is, or just substitute in values on either side of ln 2. Putting in a number just under ln 2 gives a negative; just over ln 2 gives a positive. So ln 2 must be a local minimum (well, its a global minimum too).

Can f' be undefined? (Don't forget that counts as a critical number too!) That would require e^x + 3 = 0, which is impossible. So there are no more critical numbers.

Answer 2

Well, to find the critical numbers, you must find the values for X where f'(x) = 0.. In this case such that either of the 2 terms = 0.

e^x-2 = 0
e^x = 2
x = ln 2

e^x +3 = 0
e^x = -3 ... CANNOT OCCUR, so no value of X makes this 0.

So, only one critical number... ln 2 Now, to determine if this is a min or max or neither... pick a number on the left and right of it (maybe 0 and 1) and solve for f'(x)...
f'(0) = (e^0 -2) * (e^0 +3) = (-1) * 4 = -4
f'(1) = (e^1 - 2 ) * ( e^1 + 3 ) = 0.7181.. * 5.718 = about 4
Since before ln2 the value is negative and after ln2 the value is positive, ln2 is a min-critical.

Answer 3

The critical numbers occur when the derivative is 0, so they occur here when e^x = 2 or -3. Taking the natural log means x = ln(2) or ln(-3), But you can't take the natural log of a negative number, so the only critical number is x = ln(2) or 0.63147.

I don't remember how to tell if it's a max, min or neither.

Answer 4

Critical numbers: Defined to be where f'(x) = 0 or where f'(x) = undefined.

f'(x) is 0 when the numerator is 0, so we equate it to 0.

e^x - 2 = 0
e^x = 2
Therefore, x = ln2

f'(x) is undefined when the denominator is undefined.
e^x + 3 = 0
e^x = -3, which has no solution.

Therefore, x = ln2 is a critical number.

To determine the behaviour, we test one value smaller than the critical number, and one number bigger, and plug it back into the given derivative to determine if it's positive or negative.

For the number smaller than ln2, let's choose x to be 0. Then f'(0) = (1-2)/(1+3) = (-1)/(4) = -1/4 which is a negative number, therefore, f is decreasing on (-infinity, ln2]

For the number to test bigger than ln2, let's choose x to be 1 million. Then we get ([really big number] - 2)/(really big number)+3 which is a positive number. So f is increasing on [ln2,infinity).

This also tells us that a minimum occurs at ln2, since it is decreasing before that point and increasing after that point.

Answer 5

The people who say the critical point is a minimum are correct. Polizei has the negative and postive correct, but misinterprets them. The ciritcal point is not a maximum, as Polizei says.

Answer 6

f '(x) = 0 iff x = ln2. It's easy to check that f '(x) < 0 for x < ln2 and f '(x) > 0 for x > ln2. Therefore f has a (gobal) maximum at x = ln2.