MHB Local Extrema of Quartic Function: Help Find (x, y) Points

AI Thread Summary
The discussion focuses on finding the local extrema of the quartic function f(x) = 2x^4 - 4x^2 + 6. The critical points are determined by setting the first derivative, f'(x) = 8x^3 - 8x, to zero, yielding x = -1, 0, and 1. The analysis reveals that there are relative minima at (-1, 4) and (1, 4), while the relative maximum occurs at (0, 6). The behavior of the function is confirmed through the first derivative test, indicating where the function is increasing and decreasing. The findings provide the required local maximum and minimum points for the function.
MarkFL
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Here is the question:

Can someone please help me find local maximum and local minimum?

Let f(x) = 2x^4 - 4x^2 + 6

Find the point(s) (x, y) at which f achieves:

local maximum - (?,?)
local minimum - (?,?)

can someone help me please
I get 1 and -1 but its wrong apparently

I have posted a link there to this thread so the OP can view my work.
 
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Hello John,

We are given the function:

$$f(x)=2x^4-4x^2+6$$

And are asked to find the local extrema. In order to do this, we need to equate the first derivative to zero, and finct the critical values:

$$f'(x)=8x^3-8x=8x\left(x^2-1 \right)=8x(x+1)(x-1)=0$$

Hence, our critical values are:

$$x=-1,0,1$$

Observing that the roots here are all off multiplicity 1, and seeing that for $1<x$, we have:

$$f'(x)>0$$

we may then conclude that the sign of the derivative alternates across the 4 intervals made by dividing the real number line at the three critical values, hence we have:

$$f(x)$$ increasing on $$(-1,0)\,\cup\,(1,\infty)$$

$$f(x)$$ decreasing on $$(-\infty,-1)\,\cup\,(0,1)$$

Thus, by the first derivative test, we are led to conclude that we have the following:

Relative minima at $$\left(-1,f(-1) \right)=(-1,4)$$ and $$\left(1,f(1) \right)=(1,4)$$

Relative maximum at $$\left(0,f(0) \right)=(0,6)$$

Here is a plot of the function and its relative extrema:

View attachment 2308
 

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