What is the extreme f(x)=7-(y)e^{-0.45x} ?

The extreme f(x)=7-(y)e^{-0.45x} is None

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extreme f(x)=7-(y)e^{-0.45x}

extreme points

f (x) = 7 - (y) e^{- 0.45 x}

None

Solution steps

Find the critical points:

None

7 - y e^{- 0.45 x}

has no critical points, therefore there are no extreme points

None

f (x) = x^{3} - 36 x^{2}, - 12 \leq x \leq 36

f (x) = (6 - 2 x) (3 - 2 x) x

\frac{x ^{3} + 30 x + 128}{x}, 10 \leq x \leq 20

e^{\frac{y ^{2}}{4} - x^{2} - 1}

f (x) = 6 x^{2} - 4 x + 1, 0 < x < 8

What is the extreme f(x)=7-(y)e^{-0.45x} ?
The extreme f(x)=7-(y)e^{-0.45x} is None