Question

The function g(x)=1x2-4x+4 is concave upwards on the intervals (-infinity, a) and (b, infinity) where a=? and b=?

Answer 1

that should be 1/(x^2-4x+4), sorry!

Answer 2

is this concavity problem?

Answer 3

Yes

Answer 4

\[\left(\begin{matrix}1\\ x^2-4x+4\end{matrix}\right) \]
is this the correct format?

Answer 5

finding concavity is a matter of evaluating the second derivative of a function, so
g(x)=1/(x^2-4x+4)=(x^2-4x+4)^(-1)
g''(x)=(-1)(2x-4)/(x-2)^2=-2/(x-2)=0
so we have a critical point at x=2
For x<2 like x=0
g''(0)=1>0 (concave up)
For x>2 like x=4
g''(0)=-1/2<0 (concave down)
so the interval (-infty,2) is concave down,
while (2,infty) is concave up

Answer 6

Oh okay. The fraction is what confused me.

Answer 7

I actually made a slight mistake in the derivative, it should be:

g(x)=1/(x^2-4x+4)=(x^2-4x+4)^(-1)
g''(x)=(-1)(2x-4)/(x-2)^4=-2/(x-2)^3=0
so we have a critical point at x=2
For x<2 like x=0
g''(0)=1/8>0 (concave up)
For x>2 like x=4
g''(4)=-1/8<0 (concave down)
so the interval (-infty,2) is concave down,
while (2,infty) is concave up

so the answer doesn't change

Answer 8

oh I'm sorry that's the first derivative, so we need to do the same thing for the next derivative. this should illustrate the process though.

Answer 9

Thank you!!