What are the local extema of $f (x) = x^{2} - 4 x - 5$ $f(x)=x^2-4x-5$ ?

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What are the local extema of $f (x) = x^{2} - 4 x - 5$ $f(x)=x^2-4x-5$ ?

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Answer 1 · 2016-04-27T06:11:09

At $(2, - 9)$ $(2, -9)$ There is a minima.

Explanation:

Given -

$y = x^{2} - 4 x - 5$ $y=x^2-4x-5$
Find the first two derivatives
$\frac{d y}{d x} = 2 x - 4$ $dy/dx=2x-4$

Maxima and Minima is to be determined by the second derivative.

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

At $x = 2; y = 2^{2} - 4 (2) - 5$ $x=2 ; y= 2^2-4(2)-5$

$y = 4 - 8 - 5$ $y=4-8-5$
$y = 4 - 13 = - 9$ $y=4-13=-9$

Since the second derivative is greater than one.
At $(2, - 9)$ $(2, -9)$ There is a minima.

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What are the local extema of $f (x) = x^{2} - 4 x - 5$ $f(x)=x^2-4x-5$ ?

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Explanation:

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What are the local extema of f(x)=x^2-4x-5f(x)=x2−4x−5f(x)=x^2-4x-5?

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What are the local extema of $f (x) = x^{2} - 4 x - 5$ $f(x)=x^2-4x-5$ ?