What is the extreme f(x)=-e^{(a(4-x))}+1 ?

The extreme f(x)=-e^{(a(4-x))}+1 is Saddle(4,0)

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extreme f(x)=-e^{(a(4-x))}+1

extreme points

f (x) = - e^{(a (4 - x))} + 1

Saddle (4, 0)

Solution steps

Find the critical points:

(4, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = (x - 4) e^{a (- x + 4)} (- x + 4)

D (x, a) = 2 e^{2 a (- x + 4)} x a - 8 e^{2 a (- x + 4)} a - e^{2 a (- x + 4)}

Check the sign of at each critical point

Check critical point

(4, 0) :

Saddle

Saddle (4, 0)

Z (x, y) = x + y - (\frac{x \cdot y}{100})

sin (θ) - 3 cos (θ)

f (x, y) = (- x^{2} - y^{2} + 2 y - 1)

f (x, y) = 7 (\frac{1}{5} (y + 3) e^{- x y})

y = 5 x^{2} - 15 x + 3

What is the extreme f(x)=-e^{(a(4-x))}+1 ?
The extreme f(x)=-e^{(a(4-x))}+1 is Saddle(4,0)