The table of variations is a little off because you haven't set 0 and drawn under it a double line to indicate that's it's neither differentiable nor defined in that point. f' will still be negative in ]-∞;0[ and in ]0;1[ thus requiring to calculate the limits in 0^- and -∞ since it decreases in ]-∞;0[. After studying such variations, it appears that f decreases from 0^- to -∞ hence f(x)
I got x=1 by inspection.
The solution x= 1 is obvious. The function x*e -e^x has the maximum 0 at x=1 ,therfore , x=1 is the only real solution.
Greetings, Chief Sybermath pls kindly check the math problem below
Find A, B and C,
cr[cr(2) - 1] = cr(A) + cr(B) + cr(C)
Where cr means cube root
4,2,1
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The table of variations is a little off because you haven't set 0 and drawn under it a double line to indicate that's it's neither differentiable nor defined in that point.
f' will still be negative in ]-∞;0[ and in ]0;1[ thus requiring to calculate the limits in 0^- and -∞ since it decreases in ]-∞;0[.
After studying such variations, it appears that f decreases from 0^- to -∞ hence f(x)
x = 1 using Lambert W function.
By inspection x = 1 👍👍
Trivial Solution: x=1
Interesting thought --> Taylor Series
x=-W(-1/e)=1
X=1
I took the W path and got x = -W(-1/e) before realizing that -1/e = (-1)*e^(-1) so W(-1/e)=-1 and the solution was x=1... 🤦♂
😢
So obvious,boring
e^x /x = e
x / e^x = 1/e
x. e^(-x) = 1/e
-x. e^(-x) = -1/e
W(-x. e^(-x)) = W(-1/e)
-x = W( -1. e^(-1))
-x = -1
x=1