Question

I need help plz

Answer 1

Can someone plz help me find the critical points of the following function?? \[F(x) =e ^{-x+2}\]i know the derivative of that function is is following but it don’t give me the critical points \[F’(x) = -e ^{-x+2}\]

Answer 2

Well, as you can see, there are are no solutions to the equation \(F'(x) = 0\). Might this mean that \(F\) has no critical points? What do you think?

Answer 3

I’m trying to draw what i think but it won’t let me :/ gimme a second plz :)

Answer 4

There’s a vertical asymptote at y = 0 i think

Answer 5

Are you sure that \(\ln 0 = 0\)? Remember that the domain of \(\ln\) is \((0, \infty)\). In fact, if you sketch the graph of \(\ln x\), you can see that \(\ln x \to -\infty\) as \(x \to 0^+\).

Can you sketch the graph of \(\textrm{e}^{-x+2}\)? Does it ever intersect the \(x\) axis?

Answer 6

It’s 1 pfft /.-

Answer 7

Ln 0 = 1

Answer 8

Not really. I think you mean \(\ln 1 = 0\). Actually, \(\ln 0\) is undefined.

Answer 9

:O i need to rebook at the natural log laws that i wrote in my notes.. thanks master!

Answer 10

So what do u do when you have an undefined in the equation??

Answer 11

Does that mean we souldn’t Be using this method to find the zeros?

Answer 12

Am i expected to know the vertical asymptote of this function to solve this question??

Answer 13

It means that you can't really solve that. The equation \(\textrm{e}^{-x+2} = 0\) is impossible and \(F'\) has no zeroes.

Answer 14

AAAAAAhh icy now.

Answer 15

So how do we find the critical point(s) where the graph changes from increasing to decreasing???

Answer 16

You can actually see that \(F'(x) < 0\) for all \(x\in\mathbb{R}\). This means that \(F\) is always decreasing, as you can see in the sketch you sent. Hence, there is no such point.

Answer 17

Can i write the function like that???

Answer 18

I'm afraid you can't do that. (:

Answer 19

Awww :/

Answer 20

You win i guess

Answer 21

You can convice yourself by trying to substitute \(x=0\), which yields \(\textrm{e}^2 = -\dfrac{\textrm{e}}{2}\). This is clearly false.

Answer 22

I see thanks master!! *BOWS* :)

Answer 23

I think you forgot the \(-\) sign there. \(F'(-1) = -\textrm{e}^{-(-2)+2} = -\textrm{e}^4 \simeq -54.6\), which is negative. The function \(F'\) has no zeroes at all.