Question

If g(x) = ..... and x=0 is a part of maxima of y=f(x) then

Answer 1

1)x=0 is max for g(x)
2)x=0 is min for g(x)
3)x=0 is inflexion for g(x)
4)Nothing can be said.

Answer 2

\[\Huge g(x)=3f(x)^2-2f(x)^3-6f(x)\]

Answer 3

@amistre64 @dan815 @terenzreignz

Answer 4

something tells me it's the 4th...
way too many possibilities...

Answer 5

umm no..

Answer 6

It isn't? interesting :D

Answer 7

@hartnn

Answer 8

Let's just differentiate, I suppose :3
\[\large g'(x) =\left[f(x) -f(x)^2-1\right]f'(x)\]

Answer 9

yup i got that,we also have 6 outside

Answer 10

Whoops
\[\large g'(x) =6\left[f(x) -f(x)^2-1\right]f'(x)\]

Answer 11

And differentiate again? (this'll be a nightmare)

Answer 12

\[\Large \frac{g''(x)}6 =\left[1-2f(x)\right]\left[f'(x)\right]^2 + \left[f(x) -f(x)^2-1\right]f''(x) \]

Answer 13

..

Answer 14

well if you observe if f'(x) that thing's D is <0..

Answer 15

Well, obviously, g'(0) is 0, so it's a possible extrema

Answer 16

because f'(0) is 0

Answer 17

So we need g''(0)

Answer 18

\[\Large \frac{g''(0)}6 =\left[1-2f(0)\right]\left[f'(0)\right]^2 + \left[f(0) -f(0)^2-1\right]f''(0)\]

Answer 19

This part is zero (since f'(0) is zero)
\[\Large \frac{g''(0)}6 =\color{red}{\left[1-2f(0)\right]\left[f'(0)\right]^2 }+ \left[f(0) -f(0)^2-1\right]f''(0)\]

Answer 20

So...
\[\Large \frac{g''(0)}6 =\left[f(0) -f(0)^2-1\right]f''(0)\]

Answer 21

This is negative
\[\Large \frac{g''(0)}6 =\left[f(0) -f(0)^2-1\right]\color{blue}{f''(0)}\]
since 0 is a maximum of f, second derivative at that point is negative...

Answer 22

And this part \[\Large \frac{g''(0)}6 =\color{blue}{\left[f(0) -f(0)^2-1\right]}f''(0)\]
is also negative, for whatever f(0) is...


You can check the graph of \[\Large x - x^2 - 1\]

It's all negative for whatever x is.

Answer 23

Now, that means
\[\Large \frac{g''(0)}{6}\] is positive meaning
\[\Large g''(0)\] is positive....

Answer 24

\[\Large g'(0)=0\]\[\Large g''(0)>0\]

Means 0 is a minimum of g


Case closed :P