How do you find all local maximum and minimum points given $y=x^2-98x+4$ ?

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Answer 1 · 2017-08-30T12:23:15

The minimum point is at $(49, -2397)$ [Ans]

$y= x^2-98x+4$ . At turning point , $dy/dx=0$

$dy/dx= 2x-98 :. 2x -98=0 or 2x =98 or x =49$

At $x=49 ; y = 49^2 -98*49+4 or y= -2397$

So turrning point is at $(49, -2397)$ .

$(d^2y)/dx^2= 2$ . To distinguish maximum ar minimum point

we know if $(d^2y)/dx^2 > 0$ then the point must be a minimum.

Here $(d^2y)/dx^2 > 0$ , so the point $(49, -2397)$ is minimum.

The minimum point is at $(49, -2397)$ [Ans]

How do you find all local maximum and minimum points given y=x^2-98x+4y=x^2-98x+4?