`f(x) = x^4 - 2x^2 + 3` (a) Find the intervals on which `f` is increasing or decreasing. (b) Find the local maximum and minimum values of `f`. (c) Find the intervals of concavity and the...

`f(x) = x^4 - 2x^2 + 3` (a) Find the intervals on which `f` is increasing or decreasing. (b) Find the local maximum and minimum values of `f`. (c) Find the intervals of concavity and the inflection points.

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`f(x) =x^4 - 2x^2 + 3`

(a) To solve, take the derivative of the given function.

`f'(x) =4x^3 - 4x`

Then, set the derivative equal to zero.

`0=4x^3-4x`

Factor the right side of the equation.

`0=4x(x^2-1)`

`0=4x(x-1)(x+1)`

Set each factor equal to zero. And isolate the x in each equation.

Factor 1:

`4x=0`

`x=0`

Factor 2:

`x-1=0`

`x=1`

Factor 3:

`x + 1 =0`

`x=-1`

So, f(x) has three critical numbers. These are x=-1, x=0 and x=1. The intervals formed by these three critical numbers are:

`(-oo, -1)`       `(-1,0)`        `(0,1)`    and   `(1,oo)` .

To determine which among the intervals is the function...

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`f'(x) = 4x^3 - 4x `
Let's find where `f'(x) = 0` :
`4x^3 - 4x = 0 `
`x = 0 or x = -1 or x = 1 `
If `x lt -1` , then `f'(x) lt 0` . For example, if `x = -2` , then `f'(x) = -24` . Therefore, the function is decreasing in that region.
If `-1 lt x lt 0` , then `f'(x) gt 0` . For example, if `x = -(1/2)` , then `f'(x) = 3/2` . Therefore, the function is increasing in that region.
If `0 lt x lt 1` , then `f'(x) lt 0` . For example, if `x = 1/2` , then `f'(x) = -3/2` . Therefore, the function is decreasing in that region.
If `x gt 1` , then `f'(x) gt 0` . For example, if `x = 2` , then `f'(x) = 24` . Therefore, the function is increasing in that region.
So the answer is: `f(x)` is increasing where `-1 lt x lt 0` and where `x gt 1`

`f''(x)=12x^2 - 4` and `f''(x)=0` when x = `+-1/3` .
So, f is concave up as x belongs to `(-oo,-1/3)` or `(1/3,oo) `
and f is concave down as x belongs to `(-1/3,1/3)` .

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