Question

Explain why the function g(x) = x|x| has and inflection point at (0, 0) even though g"(0) does not exist.

I know that g' is 0, and there is no rate of change for it and that's why g" DNE. But i can't see why it has an inflection point.

Answer 1

$$
\Large g(x) = x\cdot |x|
$$

Answer 2

Yes it is.

Answer 3

note that
$$
\\\Large |x|'= \frac {x}{|x|}
$$

Answer 4

Oh so looking at g'(x) we can see that there is a inflection point at x=0 if you set the numerator to zero?

Answer 5

well you need to find the second derivative, and then find where g ' ' (x) = 0 or is undefined . we see that it is undefined at x = 0.
now this is a potential inflection point.
to show it is a true inflection point , show that the concavity changes.
that meanst f ' ' (x) changes sign about x = 0

Answer 6

you can use quotient rule to find g ' ' (x)

Answer 7

$$

\Large { g(x)= x \cdot |x| \\
\text {product rule}
\\
\Large g ' (x) = 1 \cdot |x| + x \cdot \frac {x}{|x|} =
\frac{ |x|\cdot |x| }{|x|} + x \cdot \frac {x}{|x|}\\=
\frac{x^2+x^2}{|x|}= \frac{2x^2}{|x|}\\
\\ \therefore \\
g '' (x) = (\frac{2x^2}{|x|}) ' = \frac{|x|\cdot 4x -2x^2\cdot \frac{x}{|x|} }{|x|^2} = \frac {2x}{|x|}
}
$$

Answer 8

So i can prove that there is an infliction by actually finding it?

Answer 9

yes, we can actually find the inflection point.
the inflection point is where g ' ' (x) = 0 , or is undefined. (assuming this x is in the domain of g(x))

Answer 10

and g(x) is defined for x = 0.
g(0) = 0 * |0 | = 0

Answer 11

now to prove this is a true inflection point, test g ' ' ( -1) and g ' ' ( 1) .
if the sign of the second derivative changes, then x = 0 is a true inflection point

Answer 12

I got an answer that shows it does change. By looking at this how can i tell if it is concave up or down?

Answer 13

for -1 i got a negative 2nd derivative and for +1 i got a positive deriv, does that mean it is concave up?

Answer 14

any time it doesn't exist then the inflection point is (0,0)