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]

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