Question

Show that y = 1/x is concave up for x > 0 and concave down for x < 0

Answer 1

Would I start with:
y = x^-1
y' = x^-2
y'' x ^-3?

Answer 2

just sketch it...

Answer 3

or take the lim as x--->pi 0\[\lim_{x \rightarrow o^{+}} \frac{ 1 }{ x } = + \infty \]

\[\lim_{x \rightarrow 0^{-}} \frac{ 1 }{ x } = -\infty\]

Answer 4

y ' ' = -x^(-2)

y ' ' = 2x^(-3) = 2/(x^3)

It is clear when x < 0, 2/x^3 is negative and when x > 0, 2/(x^3) is a positive number.

Therefore, it is concave up when x > 0, and concave down when x < 0


Keep in mind that if y = x^(-1), then y ' ' = -x^(-2)

Answer 5

Thanks!

Answer 6

Is it clear?

Answer 7

Yes

Answer 8

ok