Question

Can someone rescue me from this problem?? I’m sick of it :/

Answer 1

Find the zeros of the following function \[F(x)= e ^{-x+2}\] its derivative is \[F’(x) = -e ^{-x+2}\]....how do i find the critical points using the derivative???

Answer 2

@Hero

Answer 3

How do i show my work to prove that f’(x)<0?????

Answer 4

I have proved that the function don’t have zeros

Answer 5

@JSVSL7 @mhchen

Answer 6

https://study.com/academy/lesson/finding-critical-points-in-calculus-function-graph-quiz.html

Read this. Maybe it can help you

Answer 7

This function is really interesting...I'm not even sure if it can be 0.
\[0=-e^{-x+2}\]

Answer 8

Is there an option for 'no solution'

Answer 9

Ummmm....all i’m Looking for are the points at which the graph changes from negative to positive but since the function have no zeros i need to show that f’(x) is less than zero which means that the curve of the function is always decreasing... sooo yeah :/

Answer 10

I guess all i’m Trying prove is that f’(x) is less than zero.. that’s all i need

Answer 11

Oh and also why’s there a vertical asymptote at x = 0....why won’t the graph cross the x-axis line

Answer 12

it did cross it.
e^x works for all real numbers

Answer 13

Bro there’s a horizontal asymptote at x = 0.... both functions don’t cross the x=axis

Answer 14

How do i algebraically figure out the asymptote ? :/

Answer 15

You can show \(y = 0\) is an asymptote of \(F'\) as \(x \to \infty\), by showing that \(\lim\limits_{x \to \infty} F'(x) = 0\).

However, your problem is finding the critical points of \(F\), which you can do by computing the zeroes of \(F'\). But \(F'\) has to zeroes, as you've seen in your last question. Hence, \(F\) has no critical points.

Answer 16

I need to show my work to prove that f’(x) is less than 0

Answer 17

How do i do that ?

Answer 18

I can’t figure out one way to prove that the 2st prime is less than 0....

Answer 19

1st prime**

Answer 20

All i wanna know is how do i prove that is 1st prime is less than 0 and the curve of the function is always decreasing???

Answer 21

Literally made a chat up there to show that some values of x do go over 0 tho..unless i’m Doing something wrong...

Answer 22

-e = -2.7182818284 right???

Answer 23

You can simply say that \(\textrm{e}^x > 0\) for all \(x \in \mathbb{R}\), so \(\textrm{e}^{-x+2} > 0\) for all \(x \in \mathbb{R}\). Then ir follows that \(F'(x) = -\textrm{e}^{-x+2} < 0\) for all \(x \in \mathbb{R}\).

If you prefer a more sophisticated argument, you can also use the Intermediate Value Theorem if you know it.

First, notice that \(F'\) is continuous in \(\mathbb{R}\). If it were to change signs within an interval \([a,b] \subset \mathbb{R}\), then it would have to cross the \(x\)-axis in order to do so, due to Intermediate Value Theorem.

But you already proved that \(F'\) has no zeroes. Hence, it is either positive or negative for all \(x \in \mathbb{R}\). So you can choose an arbitrary point, say \(x=2\) and show that \(F'(2) < 0\). Hence, \(F'(x) < 0\) for all \(x \in \mathbb{R}\).

Answer 24

I'm not quite sure what you did in your last drawing. If you're evaluating the limit as \(x \to \infty\), you can't set \(x = -2\) and then take the limit.

Answer 25

How do i do it then ?

Answer 26

Please tell me how do i show algebraically that F’(2) is less than 0??

Answer 27

oh sorry, earlier I thought when you said x=0, you meant the vertical line.

Answer 28

Well, you can simply compute it:
\(F'(2) = -\textrm{e}^{-2+2} = -\textrm{e}^0 = -1 < 0\).

Answer 29

\[0=-e^{-x+2}\]

I know you can't take natural log of both sides since ln(0) is undefined.

I think in general, \[a^{x}\] will never be 0 if a is a positive number. I'd write that on your paper

Answer 30

I agree with @mhchen above.

Answer 31

SOOOO since \[e^{-x+2}\] is always positive, then
\[-e^{-x+2}\] is always negative since positive * negative is negative

Answer 32

I was making a mistake in my algebra... i see now.. but how do i prove that there’s a horizontal asymptotes at y = 0

Answer 33

How do I algebraically prove that there’s a horizontal asymptote

Answer 34

You could find the limit as x approaches infinity or negative infinity.

The horizontal asymptote of the first derivative has nothing to do with critical points though
consider this:


sin(x)/x has a horizontal asymptote at y=0
(since as x approaches infinity, sin(x)/x approaches 0 right?)

But it also crosses the line

Answer 35

I get that the HA has nothing to do with the critical points but when i need to draw the graph based on a given function...i mostly algebraically show that there’s a Vertical or horizontal asymptote..

Answer 36

o xd, then ye just find the limit of infinity and negative infinity

Answer 37

I think Duarte gave me the answer earlier...sorry i didn’t see it....

Answer 38

\(\color{#0cbb34}{\text{Originally Posted by}}\) @DuarteME
You can show \(y = 0\) is an asymptote of \(F'\) as \(x \to \infty\), by showing that \(\lim\limits_{x \to \infty} F'(x) = 0\).

However, your problem is finding the critical points of \(F\), which you can do by computing the zeroes of \(F'\). But \(F'\) has to zeroes, as you've seen in your last question. Hence, \(F\) has no critical points.
\(\color{#0cbb34}{\text{End of Quote}}\)
Did i do something wrong there?

Answer 39

He said that F' has no zeroes, hence F has no critical points

Answer 40

\(\color{#0cbb34}{\text{Originally Posted by}}\) @DuarteME
You can simply say that \(\textrm{e}^x > 0\) for all \(x \in \mathbb{R}\), so \(\textrm{e}^{-x+2} > 0\) for all \(x \in \mathbb{R}\). Then ir follows that \(F'(x) = -\textrm{e}^{-x+2} < 0\) for all \(x \in \mathbb{R}\).

\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 41

I could do that but i wanna know how do i show that \[\lim_{x \rightarrow \infty} F’(x) = 0 \]

Answer 42

Actually I figured it out... lol ....sorry guys but it’s one of those tedious questions...

Answer 43

lol I'm too sleepy now

Answer 44

Sorry mate.. *~*

Answer 45

Thanks guys!! I promise the rest of the questions will be much more straight forwards lol

Answer 46

I’ll post my solutions plz check them :)

Answer 47

God bless y’all!!