Question

Need help with the following questions:

Answer 1

1) If f(x)=x^4-14x^3+72X^2+9x-10
find the largest interval on which f(x) is concave downward. If we write the interval as (a,b), then a=___ and b=___

2) If g(x)=x^4-12x^3+30x^2+9x-3
find the largest interval on which g(x) is concave upward. If we write the interval as (-infinity, a) U (b, infinity), then a=___ and b=___

3) If f(x)=xe^6x
find the largest interval on which f(x) is concave upward. If we write the interval as (a, infinity), then a=___

Answer 2

do i take the derivative of each twice to get these answers?

Answer 3

@satellite73
@amistre64
@KingGeorge
@.Sam.
@Mertsj
@saifoo.khan

Answer 4

someone help? i dont understand this stuff completely yet

Answer 5

You would take the second derivative twice, and find the zeros to determine where it switches concavity.

Answer 6

Then plug in values in the different intervals you get to find if the concavity is positive or negative in that interval.

Answer 7

show me the work for one? i get the gist of it, just wanna make sure

Answer 8

Let's take the first one then. You have\[f(x)=x^4-14x^3+72x^2+9x-10\]so\[f'(x)=4x^3-42x^2+144x+9\]\[f''(x)=12x^2-84x+144=12(x^2-7x+12)\]Now you set that equal to zero, and you get\[0=12(x^2-7x+12)=(x-3)(x-4)\]So you have zeros at \(x=3\), \(x=4\).

Answer 9

Finally, plug in some number for \(x\) that is less than three, one that is between 3 and 4, and one that is greater than 4, and check which one(s) are negative. In this case, only the one between 3 and 4 is negative, so that must be greatest interval on which \(f(x)\) is concave downward.

Answer 10

what happened to the 12 up there. did you divide by 12 on both sides?

Answer 11

Yup.

Answer 12

so the answer would be like 3.5

Answer 13

You want an interval, not a single number, so your interval should be \((3, 4)\). Be careful that it doesn't include \(3\) or \(4\), since the concavity is zero there.

Answer 14

o yeah duh. i wasnt thinking straight.

Answer 15

when we plug the numbers less, in between, and greater than, we plug them in the original function, correct?

Answer 16

Not if you want to find the concavity. You would want to plug them into the second derivative.

Answer 17

o ok

Answer 18

i know prob 3 is similar but could u help me with that as well...

Answer 19

3) If f(x)=xe^6x
find the largest interval on which f(x) is concave upward. If we write the interval as (a, infinity), then a=___

Answer 20

First we need to find the second derivative. Using the product rule and chain rule\[f(x)=xe^{6x}\]\[f'(x)=6xe^{6x} + e^{6x}=e^{6x}(6x+1)\]\[f''(x)=6e^{6x}+36xe^{6x}\]Now you find the zeros.

Answer 21

So\[0=6e^{6x}+36xe^{6x}=e^{6x}+6xe^{6x}=e^{6x}(1+6x)\] since \(e^{6x}\) is never 0, the only zero is at \(1+6x=0 \to x=-1/6\)

Answer 22

so whats the final answer im confused

Answer 23

calc is so complex :(

Answer 24

That's where the concavity switches. So if we plug in \(x=0\), we have that the concavity is upward from \((-1/6, \infty)\)

Answer 25

i put that as my answer and it said it was wrong

Answer 26

u sure u did it right?

Answer 27

im confused by some of ur work above ^^^

Answer 28

You have to notice that the question is calling for a single value \(a\), not an interval. Since it gives you the interval as \((a, \infty)\), your answer should be \(a=-1/6\).

Answer 29

i know. it wont accept that

Answer 30

I think I did the second derivative wrong. It should actually be\[f''(x)=12e^{6x}(3x+1)\]

Answer 31

So \(a=-1/3\) instead.

Answer 32

ok thats its. can u rewrite it all correctly?

Answer 33

*it

Answer 34

Give me a minute.

Answer 35

sure

Answer 36

\[f(x)=xe^{6x}\]\[f'(x)=6 xe^{6x} +e^{6x} = e^{6x}(6x+1)\]\[f''(x)=6 e^{6x} (6 x+1)+6 e^{6x}=12e^{6x}(3x+1)\]Since \(12e^{6x}\) is never zero, we know that \((3x+1)=0\) So \(x=-1/3\).

Answer 37

THANKS

Answer 38

You're welcome. Sorry it took so long.

Answer 39

its fine. been doing other problems while waiting.