What is the extreme f(x)=e^x,0<= x<= ln(18) ?

The extreme f(x)=e^x,0<= x<= ln(18) is Minimum(0,1),Maximum(ln(18),18)

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extreme f(x)=e^x,0<= x<= ln(18)

extreme points

f (x) = e^{x}, 0 \leq x \leq ln (18)

Minimum (0, 1), Maximum (ln (18), 18)

Solution steps

Find the critical points:

x = 0, x = ln (18)

Domain of

e^{x}, 0 \leq x \leq ln (18) : 0 \leq x \leq ln (18)

Find intervals:

Increasing

: 0 < x < ln (18)

Plug the extreme point

x = 0

into

e^{x} \Rightarrow y = 1

Minimum (0, 1)

Plug the extreme point

x = ln (18)

into

e^{x} \Rightarrow y = 18

Maximum (ln (18), 18)

Minimum (0, 1), Maximum (ln (18), 18)

f (x) = 2 x^{3} - 6 x^{2} - 210 x + 7

f (x) = - 3 x^{5} + 5 x^{3} + 1

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

f (x, y) = x^{2} + y^{2} + 4 x - 2 y + 6

f (x) = \frac{4860}{x} + 18 x + 724924

What is the extreme f(x)=e^x,0<= x<= ln(18) ?
The extreme f(x)=e^x,0<= x<= ln(18) is Minimum(0,1),Maximum(ln(18),18)