For which val's of c does f(x) = cx^4 - 2x^2 + 1 have max pt

[grapz]

f(x) = c x^4 - 2x^2 + 1

For which values of c do the curves have maximum points?

I found the first derivative

4c x^ 3 - 4x

and factored

4x ( cx^2 - 1)

x = 0, 1/ c^1/2

I am not sure what to do next :/

 

[Shriya]

wat is the domain of x???

 

[soroban]

Re: For which val's of c does f(x) = cx^4 - 2x^2 + 1 have ma

Hello, grapz!

\(\displaystyle f(x) \:=\:c x^4\,-\,2x^2\,+\,1\)

For which values of \(\displaystyle c\) do the curves have maximum points?

I found the first derivative: \(\displaystyle \:4cx^3\,-\,4x\:=\:0\)
and factored: \(\displaystyle \:4x(cx^2\,-\,1) \:=\:0\;\;\Rightarrow\;\;x \:= \:0,\:\L\pm\frac{1}{\sqrt{c}}\;\) . . . Good!

We see immediately that: \(\displaystyle \,c\,>\,0\)

Now test your critical values and see if we get a maximum.


Second derivative: \(\displaystyle \:f''(x)\:=\:12cx^2\,-\,4\)

At \(\displaystyle x\,=\,\pm\frac{1}{\sqrt{c}}:\;f''(x) \:=\:12c\left(\pm\frac{1}{\sqrt{c}}\right)^2\,-\,4\:=\:12c\left(\frac{1}{c}\right)\,-\,4 \:=\:+8\)
. . Concave up: \(\displaystyle \cup\) . . . minimum

At \(\displaystyle x\,=\,0:\;f''(x)\:=\:12c(0^2)\,-\,4\:=\:-4\)
. . Concave down: \(\displaystyle \cap\) . . . maximum


Bottom line: \(\displaystyle f(x)\) has a maximum for any \(\displaystyle c\:>\:0.\)