What is the extreme f(x)=tan(x),-pi/6 <= x<= pi/4 ?

The extreme f(x)=tan(x),-pi/6 <= x<= pi/4 is Minimum(-pi/6 ,-(sqrt(3))/3),Maximum(pi/4 ,1)

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extreme f(x)=tan(x),-pi/6 <= x<= pi/4

extreme points

f (x) = tan (x), - \frac{π}{6} \leq x \leq \frac{π}{4}

Minimum (- \frac{π}{6}, - \frac{3}{3}), Maximum (\frac{π}{4}, 1)

Solution steps

Find the critical points:

x = - \frac{π}{6}, x = \frac{π}{4}

Domain of

tan (x), - \frac{π}{6} \leq x \leq \frac{π}{4} : - \frac{π}{6} \leq x \leq \frac{π}{4}

Find intervals:

Increasing

: - \frac{π}{6} < x < \frac{π}{4}

Plug the extreme point

x = - \frac{π}{6}

into

tan (x) \Rightarrow y = - \frac{3}{3}

Minimum (- \frac{π}{6}, - \frac{3}{3})

Plug the extreme point

x = \frac{π}{4}

into

tan (x) \Rightarrow y = 1

Maximum (\frac{π}{4}, 1)

Minimum (- \frac{π}{6}, - \frac{3}{3}), Maximum (\frac{π}{4}, 1)

f (x) = x^{4} - 4 x^{3} - 8

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

f (x) = x^{\frac{2}{5}} (3 x + 7)

x^{3} - 2 x^{2} - 15 x + 10 [- 2.1]

f (x) = x + \frac{9}{x} + 2

What is the extreme f(x)=tan(x),-pi/6 <= x<= pi/4 ?
The extreme f(x)=tan(x),-pi/6 <= x<= pi/4 is Minimum(-pi/6 ,-(sqrt(3))/3),Maximum(pi/4 ,1)