# How many inflection points does (tan^2)x have on the interval (-pi,2pi)

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### 1 Answer

`y = tan^2(x)`

To find the extreme points we must first find the x values on which y' is zero and then check for y'' on those points to determine whether they are inflection points.

`y' = 2tan(x)sec^2(x)`

`y' = 2tan(x)(tan^2(x)+1)`

For extreme points y' = 0,

This can only happen if `tan(x) = 0` .

**In the given range the solutions for** `tan(x) = 0` are `-pi, 0, pi ` and `2pi`

Now let's find `y''` . For inflection points, `y'' = 0 .`

`y'' = 2tan(x)xx2sec(x)xxsec(x)tan(x) + 2sec^2(x)sec^2(x)`

`y'' = 4sec^2(x)tan^2(x) + 2sec^4(x)`

`y'' = 2sec^2(x)(2tan^2(x)+sec^2(x))`

**This is always greater than zero. For any x, y'' > 0. Therefore, there are no inflection points in the given range**