How do you find the maximum value of #f(x) = -3x^2 +30x -50#?

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Answer 1 · 2017-08-26T11:35:14

Maximum value of # f(x) # is #25#

#f (x) = -3x^2 +30x -50 #

#f^' (x) = -6x +30 # , At stationary point #f^' (x) = 0# or

#-6x +30 = 0 or x = 30/6 =5 #

At #x=5 ; f(x) = - 3(5^2)+30*(5) -50= -75+150-50=25#

Stationary point is # (5, 25) #

2nd derivative test for maximum or minimum :

# (d^2y)/dx^2 = -6 # (negative)

If #(d^2y)/dx^2# is negative, then the point is a maximum turning point.

So #(5,25) # is maximum turning point and maximum value of

# f(x) # is #25#

graph{-3x^2+30x-50 [-50.64, 50.6, -25.28, 25.37]} [Ans]