Question

Find the CONCAVE DOWN and CONCAVE UP on the function f(x)=e^x - x^2 by setting the 2nd Derivative equal to 0!!

Answer 1

First D= e^x - 2x

Second D= e^x - 2

Set equal to 0--> e^x - 2=0

Now what??

Answer 2

Hmm let's add 2 to each side maybe? :)

Answer 3

\[\large\rm e^x=2\]Now we need to get that x out of the exponent some how...
hmmm

Answer 4

Remember anything about logs?

Answer 5

a bit... so would we bring the x in front of the e?

Answer 6

Log is the inverse operation of the exponential.
When we take the composition of inverses,
they essentially "undo" one another.
\[\large\rm \ln(e^x)=x\ln(e)=x\]

In the same way you would undo multiplication with its inverse (division),
or undo subtraction with its inverse (addition, like we did earlier in this problem),
you can undo exponentiation with its inverse (log).

Answer 7

So we'll apply natural log to each side,\[\large\rm \ln(e^x)=\ln(2)\]

Answer 8

\[\large\rm x=\ln(2)\]And that gives you your inflection point, ya? :)

Answer 9

my inflection point?

Answer 10

what is an inflection point? i am trying to find where the concave up and the concave down are located... i need to know if it is a positive or negative 2nd derivative

Answer 11

Inflection point is where the function changes concavity :)

Answer 12

Second derivative is ZERO at x=ln(2).
That is the location where it changes from negative to positive! :)

Answer 13

(ln(2),f(ln(2))) is a possible inflection point of f

it is an inflection point if the concavity switches at that point

you can choose test numbers from before and after ln(2) to check to see if (ln(2),f(ln(2)) is an inflection point (plug into f'' because that is what tells you if the function f is concave up/down)
Concave up means f''>0
Concave down means f''<0

So you are testing the following two intervals:
(-inf,ln(2))
and
(ln(2),inf)
by choosing a number from each interval and pluggin into f''

Answer 14

Basically,
\(\color{#000000 }{ \displaystyle f''(\ln 2 \pm \delta) }\) have to have different signs.

where \(\delta\) is a sufficiently close value (like 0.3 or 1 even).

Answer 15

have to have different signs, for \(x=ln(2)\) to be indeed the reflection point.

Answer 16

:p inflection point

Answer 17

Oh, lol, yes \(\color{red}{\rm in}\)flection point.

Answer 18

Amenah ?

Answer 19

sorry! i was doing other homework :)

Answer 20

ok, so what can you say about this problem so far?

Do you have questions about it, you understand it or.....

Answer 21

how did you know the second derivative was 0 at x=ln(2)?

Answer 22

You have found that the second derivative was,

Example Problem:
\(\color{#000000 }{ \displaystyle f''(x)=e^x-2 }\)

Right

Answer 23

right

Answer 24

should have "example problem" there

Answer 25

\(\color{#000000 }{ \displaystyle f''(x)=e^x-2 }\)

you want to know at what value(s) of x is your second derivative is 0?
So you set

\(\color{#000000 }{ \displaystyle 0=e^x-2 }\)

Answer 26

Then, you solve for x.

Answer 27

and we did that by doing: ln(e^x) = ln2

Answer 28

how would i use that to tell where the concave up/down is? is this right: x<ln(2)
x>ln(2)

Answer 29

how would i see if they are positive or negative?

Answer 30

yes, plug in some values like ln(2)+0.1 and ln(2)-0.1
into the second derivative.

Answer 31

If there are going to have different signs, then your cancavity indeed changes, and thus ln(2) is indeed an inflection point. If the sign is going to be the same, then x=ln(2) is not an inflection point.

Answer 32

so i plug points into: e^x - 2x ?

Answer 33

yes, your second derivative is
\(\color{#000000 }{ \displaystyle f''(x)=e^x-2 }\)

so you plug in
\(\color{#000000 }{ \displaystyle x=\ln(2)\pm \delta }\)

into the \(\color{#000000 }{ \displaystyle f''(x) }\)

you can choose your delta to be \(\color{#000000 }{ \displaystyle \delta =1 }\)

Answer 34

is this thing about delta too confusing?

Answer 35

yeah, we haven't learned that yet :)

Answer 36

well, I am just using that notation and name, all I want to know if the second derivative changes signs about x=ln(2)

Answer 37

Do you want an example problem

Answer 38

?

Answer 39

sorry! i'm here :) yes, i would like an example, please.

Answer 40

Find the concave down and concave up on the function
\(\color{#000000 }{ \displaystyle f(x)=xe^x }\) by setting \(\color{#000000 }{ \displaystyle f''(x)=0 }\).

We will differentiate this function through the product rule:
\(\color{#000000 }{ \displaystyle d/dx~e^x=e^x }\)
\(\color{#000000 }{ \displaystyle d/dx~x=1 }\)
(So, \(\color{#000000 }{ \displaystyle d/dx~xe^x=e^x+xe^x }\))

Therefore, we know that,
\(\color{#000000 }{ \displaystyle f'(x)=e^x+xe^x }\)

And then we will differentiate this function again to get the second derivative. (note: we already know \(\color{#000000 }{ \displaystyle d/dx~xe^x }\))

\(\color{#000000 }{ \displaystyle f''(x)=e^x+(e^x+xe^x) }\)
\(\color{#000000 }{ \displaystyle f''(x)=2e^x+xe^x }\)

We want to find possible inflection points, so we set,
\(\color{#000000 }{ \displaystyle 0=2e^x+xe^x }\)
\(\color{#000000 }{ \displaystyle 0=e^x(2+x) }\)

We know that \(\color{#000000 }{ \displaystyle e^x>0~\forall x\in\mathbb{R} }\), so dividing by \(e^x\) on both sides would have given us \(e^x=0\) as an alternative solution, but that is not a possibility since \(e^x\) is always positive for real numbers)

\(\color{#000000 }{ \displaystyle 0=2+x}\)
\(\color{#000000 }{ \displaystyle -2=x}\)

Answer 41

Now, you would like to test \(x=2\) to see if this actually an inflection point!

Answer 42

So, what you need to verify that on one sides of \(x=-2\), the second derivative is negative and on the other sides of \(x=-2\) is positive. (i.e., you want to verify the change in signs of the second derivative, which indicate the change in concavity of the function)

review:
the function is concave up at the x-values that satisfy f''(x)>0,
the function concave down at the x values that satisfy f''(x)<0.

Recall that,
\(\color{#000000 }{ \displaystyle f''(x)=2e^x+xe^x }\)


\(\color{#000000 }{ \displaystyle f''(-2+\color{blue}{1})=f''(-1)=2e^{-1}+(-1)e^{-1}=e^{-1}=\frac{1}{e}>0 }\)

\(\color{#000000 }{ \displaystyle f''(-2-\color{blue}{1})=f''(-3)=2e^{-3}+(-3)e^{-3}=-e^{1}<0 }\)

NOTE:
I choose my delta; \(\color{#000000 }{ \displaystyle \delta =1 }\)
(but choosing VERY large \(\delta\) might result in failure, however \(\delta=1\) is close enough)


So we can imply that for \(\color{#000000 }{ \displaystyle x = (-\infty~,~ -2) }\),
\(\color{#000000 }{ \displaystyle f''(x)<0 }\)
AND for \(\color{#000000 }{ \displaystyle x = (-2~,~+\infty) }\),
\(\color{#000000 }{ \displaystyle f''(x)>0 }\)

Answer 43

So since the \(f''(x)\) indeed changes signs about \(x=-2\), (as I have shown choosing \(\delta=1\)), THEREFORE \(x=-2\) is an inflcetion point.

Answer 44

And for (-2, ∞) the function is concave up (since second derivative is positive for these values),

and for (-∞, -2) the function is caoncave down (since the second derivative is negative at those values)

Answer 45

why is it -2? shouldn't it be positive?

Answer 46

so in this problem, our inflection point is ln(2)

Answer 47

in the example problem x=-2 is the inflection point. I showed how.

Answer 48

oh! i see

Answer 49

in oru problem the inflection point is ln(2), but you need to verify this, by evaluating

\(f''(\ln(2)\pm\delta)\), and proving that they have different signs.

Answer 50

what \(\delta\) do you want to choose?
(has to be sufficiently small)

Answer 51

1

Answer 52

Ok, good, so you need to evaluate
\(f''(\ln(2)\pm1)\)

Answer 53

find
f''(ln(2)+1)

and
f''(ln(2)-1)

Answer 54

can you wait a moment while i grab my calculator?

Answer 55

sure

Answer 56

1.7 & -.31

Answer 57

ok, so you know that for x>ln(2), the second derivative is positive, so the function is concave up for x>ln(2).

ok, so you know that for x<ln(2), the second derivative is negative, so the function is concave down for x<ln(2).

Answer 58

i got it! thank you so much :)

Answer 59

You welcome!