What are the extrema of $f (x) = x^{2} - 8 x + 12$ $f(x)=x^2 - 8x + 12$ on $[- 2, 4]$ $[-2,4]$ ?

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What are the extrema of $f (x) = x^{2} - 8 x + 12$ $f(x)=x^2 - 8x + 12$ on $[- 2, 4]$ $[-2,4]$ ?

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Answer 1 · 2016-08-21T04:58:02

the function has a minimum at $x = 4$ $x=4$ graph{x^2-8x+12 [-10, 10, -5, 5]}

Explanation:

Given -

$y = x^{2} - 8 x + 12$ $y=x^2-8x+12$

$\frac{d y}{d x} = 2 x - 8$ $dy/dx=2x-8$

$\frac{d y}{d x} = 0 \Rightarrow 2 x - 8 = 0$ $dy/dx=0 => 2x-8=0$

$x = \frac{8}{2} = 4$ $x=8/2=4$

$\frac{d^{2} y}{{d x}^{2}} = 2 > 0$ $(d^2y)/(dx^2)=2>0$

At $x = 4; \frac{d y}{d x} = 0; \frac{d^{2} y}{{d x}^{2}} > 0$ $x=4; dy/dx=0;(d^2y)/(dx^2)>0$

Hence the function has a minimum at $x = 4$ $x=4$

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What are the extrema of $f (x) = x^{2} - 8 x + 12$ $f(x)=x^2 - 8x + 12$ on $[- 2, 4]$ $[-2,4]$ ?

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Explanation:

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Science

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What are the extrema of f(x)=x2−8x+12f(x)=x^2 - 8x + 12 on [−2,4][-2,4]?

1 Answer

Explanation:

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What are the extrema of $f (x) = x^{2} - 8 x + 12$ $f(x)=x^2 - 8x + 12$ on $[- 2, 4]$ $[-2,4]$ ?