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]