What is the extreme f(x)=40e^{2x}-5e^x ?

The extreme f(x)=40e^{2x}-5e^x is Minimum(-4ln(2),-5/32)

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extreme f(x)=40e^{2x}-5e^x

extreme points

f (x) = 40 e^{2 x} - 5 e^{x}

Minimum (- 4 ln (2), - \frac{5}{32})

Solution steps

f^{'} (x) = 80 e^{2 x} - 5 e^{x}

Find intervals:

Decreasing

: - \infty < x < - 4 ln (2),

Increasing

: - 4 ln (2) < x < \infty

Plug

x = - 4 ln (2)

into

40 e^{2 x} - 5 e^{x} : - \frac{5}{32}

Minimum (- 4 ln (2), - \frac{5}{32})

f (x) = x + \frac{49}{x}, 2 \leq x \leq 12

f (x) = 3 - 6 x^{2}

f (x, y) = x^{2} + x y - y^{2} - y

f (x) = \frac{( x - 1 ) ^{2}}{x - 3}

f (x) = 3 x^{2} + \frac{48}{x ^{2}}

What is the extreme f(x)=40e^{2x}-5e^x ?
The extreme f(x)=40e^{2x}-5e^x is Minimum(-4ln(2),-5/32)