What is the extreme f(x)=4x-(196)/x ,1<= x<= 9 ?

The extreme f(x)=4x-(196)/x ,1<= x<= 9 is Minimum(1,-192),Maximum(9, 128/9)

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extreme f(x)=4x-(196)/x ,1<= x<= 9

extreme points

f (x) = 4 x - \frac{196}{x}, 1 \leq x \leq 9

Minimum (1, - 192), Maximum (9, \frac{128}{9})

Solution steps

Find the critical points:

x = 1, x = 9

Domain of

4 x - \frac{196}{x}, 1 \leq x \leq 9 : 1 \leq x \leq 9

Find intervals:

Increasing

: 1 < x < 9

Plug the extreme point

x = 1

into

4 x - \frac{196}{x} \Rightarrow y = - 192

Minimum (1, - 192)

Plug the extreme point

x = 9

into

4 x - \frac{196}{x} \Rightarrow y = \frac{128}{9}

Maximum (9, \frac{128}{9})

Minimum (1, - 192), Maximum (9, \frac{128}{9})

\frac{x - 1}{x ^{2} + 2 x + 6}

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

f (x) = 3 (x + 1)^{2} - 4

y = 2 x^{2} - 3

2 x + 6

What is the extreme f(x)=4x-(196)/x ,1<= x<= 9 ?
The extreme f(x)=4x-(196)/x ,1<= x<= 9 is Minimum(1,-192),Maximum(9, 128/9)