What is the extreme f(x)=-0.1t^2+1.2t+98.3 ?

The extreme f(x)=-0.1t^2+1.2t+98.3 is Maximum(6,101.9)

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extreme f(x)=-0.1t^2+1.2t+98.3

extreme points

f (x) = - 0.1 t^{2} + 1.2 t + 98.3

Maximum (6, 101.9)

Solution steps

f^{'} (t) = - 0.2 t + 1.2

Find intervals:

Increasing

: - \infty < t < 6,

Decreasing

: 6 < t < \infty

Plug

t = 6

into

- 0.1 t^{2} + 1.2 t + 98.3 : 101.9

Maximum (6, 101.9)

f (x) = 2 x x - 5

x - 1

x + 16

f (x) = (\frac{1}{3} x - 3)

f (x) = \frac{1}{2} (x - 4)^{5} + 1

What is the extreme f(x)=-0.1t^2+1.2t+98.3 ?
The extreme f(x)=-0.1t^2+1.2t+98.3 is Maximum(6,101.9)