Question

You have function f(x) = e^x + x^2 - 4 = 0
Thus, f'(x) = e^x + 2x

How can we find out algebraically where f'(x) = 0?

Answer 1

you cannot

Answer 2

You cannot even do that sub in and check LHS = RHS trick?

Answer 3

i mean you cannot using algebra

Answer 4

ie. sub in some x and see if LHS = RHS?

Answer 5

easy to check
see what the answer is here, and imagine how you would find it
http://www.wolframalpha.com/input/?i=e^x+%2B+2x+

Answer 6

What do you mean "easy to check"?

Answer 7

Oh you are saying go look it up on Wolf

Answer 8

No it has to be x < 0

Answer 9

hmmm i could tell you that x has to be greater than zero not much of a help because we got e^x = -2x and then if we ln both sides we get x= ln(-2x) and in order for ln to work it must be positive so hence -2x> 0 and that implies x<0

Answer 10

:)

Answer 11

yeah i just noticed that lol

Answer 12

Done. That is all that I needed to know: that inequality.

Answer 13

o awesome lol :D

Answer 14

i meant it is easy to check that it is hard to solve!