What is the extreme f(x)=e^xx+2,0<= x<= 6 ?

The extreme f(x)=e^xx+2,0<= x<= 6 is Minimum(0,2),Maximum(6,e^6*6+2)

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extreme f(x)=e^xx+2,0<= x<= 6

extreme points

f (x) = e^{x} x + 2, 0 \leq x \leq 6

Minimum (0, 2), Maximum (6, e^{6} \cdot 6 + 2)

Solution steps

Find the critical points:

x = 0, x = 6

Domain of

e^{x} x + 2, 0 \leq x \leq 6 : 0 \leq x \leq 6

Find intervals:

Increasing

: 0 < x < 6

Plug the extreme point

x = 0

into

e^{x} x + 2 \Rightarrow y = 2

Minimum (0, 2)

Plug the extreme point

x = 6

into

e^{x} x + 2 \Rightarrow y = e^{6} 6 + 2

Maximum (6, e^{6} \cdot 6 + 2)

Minimum (0, 2), Maximum (6, e^{6} \cdot 6 + 2)

y = \frac{27 x ^{2}}{( 2 - x ) ^{3}}

f (x, y) = 6 x + 6 y + 2 x y - 5 x^{2} - 2 y^{2}

f (x) 2 x^{2} - 4 x - 1

5 cos^{2} (x), 0 \leq x \leq π

y = x^{3} - 6 x^{2} + 15

What is the extreme f(x)=e^xx+2,0<= x<= 6 ?
The extreme f(x)=e^xx+2,0<= x<= 6 is Minimum(0,2),Maximum(6,e^6*6+2)