Question

f(x)= 7x/(x^2-9). use interval notation to indicate where f(x) is decreasing. what is the local min and max and inflection points

Answer 1

First find the derivative
\[
f'(x)=\frac {7(x^2-9) - 2x(7x) } {(x^2 -9)^2}=
\frac {-7 x^2-63 } {(x^2 -9)^2}=-\frac {7 (x^2+9) } {(x^2 -9)^2}
\]

Answer 2

What can you say about the sign of the derivative?

Answer 3

the negative sign? I'm not sure what do you mean

Answer 4

Is
\[x^2 + 9
\]
always >0?

Answer 5

no... you can't take the root of a negative number!

Answer 6

Is
\[
(x^2 -9)^2 \ge 0
\]

Answer 7

Give me a real number x such that
\[
x^2 + 9 \le 0
\]
There is not such a number.
\[
x^2 \ge 0\\
9 > 0\\
x^2 + 9 > 0
\]

Answer 8

so its none? or does not exist?

Answer 9

I mean \[ x^2 + 9\] is always strictly bigger than zero.

Answer 10

how do i find the interval that it is decreasing?

Answer 11

The derivative
\[
-\frac {7 (x^2+9) } {(x^2 -9)^2}
\]
is always < 0 when it is defined. So f is decreasing everywhere in its domain.
The domain is every x except x=-3 and x =3.

Answer 12

Did you get it?

Answer 13

ya thanks!

Answer 14

so to find out if it is concave up or down i take the derivative again right?

Answer 15

Can this function have local max or local min?

Answer 16

Yes

Answer 17

no it can not

Answer 18

great.

Answer 19

I'm supposed to find interval notation of when it is concave up and when it is concave down. so the second derivative numerator is 14x(x^2+27)... so x^2+27>0? i don't get what i do after to find the intervals

Answer 20

Now try to find the second derivative step by step. To check your answer,
\[
f''(x)=\frac{14 x \left(x^2+27
\right)}{\left(x^2-9\right
)^3}
\]
You can argue that the sign of f''(x) is the sign of 14 x. Why?

Answer 21

because then you distribute it? do you mean why it is positive?

Answer 22

yes

Answer 23

that x is greater than zero since our f(x) is positive?

Answer 24

No, that means that x=0 is an inflection point.

Answer 25

ohhh

Answer 26

There are two other points x=-3 and x=3 where f has vertical asymptotes where the
concavity changes.

Answer 27

The three points that you should worry about are
x -Infinity -3 0 3 +Infinity
f'(x) - + 0 - +
f(x) conc dn con up conc dn conc up

Answer 28

f''(x) in the second rwo

Answer 29

ohhh i get it! perfect

Answer 30

f is concave up on (-3,0) and (3, Infinity)
Where is concave down?

Answer 31

(-inf, -3) and (0,3)

Answer 32

yes

Answer 33

thank you!!!!!!!

Answer 34

yw

Answer 35

:)