find the minimum of f(x) = 2x^2 - 4x + 3

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Find the minimum by taking the derivitive of the function and setting it to zero.

Use the power rule which states: d/dx [x^n] = n*x^(n-1)

f(x) = 2x^2 - 4x^1 + 3

f'(x) = 4x -4

Now set to zero

4x - 4 = 0

x = 1

minimum is at f(1) = 2 - 4 + 3 = 1 , or the point (1,1)

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

First we need to determine f'(x)

f'(x) = 4x - 4

Now we need to find the critical values for f(x) which is f'(x)'s zero.

4x - 4 = 0

==> 4x = 4

==> x= 1

Then f'(x) has a minimmum values at x= 1

==> f(1) = 2(1) - 4(1) + 3

= 2 - 4 + 3

= 1

Then f has a minimmum value at (1, 1)

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

We have 2(x-1)^2>=0

so f(x)=2(x-1)^2+1>=1

f(x) minimum=1 when x=1

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