What is the extreme f(x)=2x+(300000)/x ,1<= x<= 300 ?

The extreme f(x)=2x+(300000)/x ,1<= x<= 300 is Maximum(1,300002),Minimum(300,1600)

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extreme f(x)=2x+(300000)/x ,1<= x<= 300

extreme points

f (x) = 2 x + \frac{300000}{x}, 1 \leq x \leq 300

Maximum (1, 300002), Minimum (300, 1600)

Solution steps

Find the critical points:

x = 1, x = 300

Domain of

2 x + \frac{300000}{x}, 1 \leq x \leq 300 : 1 \leq x \leq 300

Find intervals:

Decreasing

: 1 < x < 300

Plug the extreme point

x = 1

into

2 x + \frac{300000}{x} \Rightarrow y = 300002

Maximum (1, 300002)

Plug the extreme point

x = 300

into

2 x + \frac{300000}{x} \Rightarrow y = 1600

Minimum (300, 1600)

Maximum (1, 300002), Minimum (300, 1600)

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

f (x) = - x^{4} + 4 x^{3}

\frac{1}{3} x^{3} - x^{2} + x - 5

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

f (x) = 98 y^{2} + x^{2} - x^{2} y

What is the extreme f(x)=2x+(300000)/x ,1<= x<= 300 ?
The extreme f(x)=2x+(300000)/x ,1<= x<= 300 is Maximum(1,300002),Minimum(300,1600)