Question

Find all the intervals on which
f(x) = x^2 + 1 / x^2
is concave upward

Answer 1

1) find the second derivative
i will wait

Answer 2

ok

Answer 3

6x^-4

Answer 4

hmm no

Answer 5

the first derivative is -2x^-3

Answer 6

what happened to the \(x^2\) part? did it disparu?

Answer 7

you were close, you just forgot the \(2\)

Answer 8

what do you mean

Answer 9

-2 times -3 is 6

Answer 10

\[f(x)=\color{red}{x^2}+\frac{1}{x^2}\] right?

Answer 11

yes

Answer 12

you seem to be ignoring the poor \(\color{red}{x^2}\) part
what did it do to you?

Answer 13

I am not sure what you mean

Answer 14

i used the quotient rule

Answer 15

you gotta take the second derivative of that as well

Answer 16

you take the second derivative of each term.

Answer 17

yes using the quotient rule I got
f'(x) = -2x^-3 and
f''(x) = 6x^-4

Answer 18

https://www.mathway.com/

Answer 19

\[y=x^2\\
y'=2x\\
y''=2\]

Answer 20

you got the second derivative right for the
\[\frac{1}{x^2}\] part, but you forgot about the \(x^2\) part

Answer 21

2x-2x^-3 is the first derivative and the secon is 2+6x^-4

Answer 22

yay

Answer 23

see what i meant about the 2?

Answer 24

yeah haha

Answer 25

now it is nice to use exponential notation to find derivatives, but to compute you need positive exponents
\[f''(x)=2+\frac{6}{x^4}\]

Answer 26

and your job is to find out where this beast is positive, and where it is negative, i.e. for which values of x is it positive or negative, which we do with our eyeballs

Answer 27

how about \(x^4\) is that ever negative?

Answer 28

no

Answer 29

right

Answer 30

and therefore neither is \(\frac{6}{x^4}\)

Answer 31

and if \(\frac{6}{x^4}\) is never negative, if you add 2 it will not be negative either (remember that 2?)

Answer 32

yeah

Answer 33

so the second derivative is always positive, (unless it its undefined)

Answer 34

making your original function
\[f(x)=x^2+\frac{1}{x^2}\] always concave _________

Answer 35

which is the original