How do you find all local maximum and minimum points using the second derivative test given #y=tan^2x#?

Question

Christopher Agresta · Accepted Answer

graph{(tanx)^2 [-7.09, 12.91, -2.12, 7.88]}

It has minimum at #x=kpi, kinZZ# valued 0

#y'=2(tanx)/cos^2x# #y'=0, AAx=kpi, kinZZ# #y''=2*(1/cos^2x*cos^2x+2tanxcosxsinx)/(cos^4x)=2*(1+2sin^2x)/cos^4x>0, AAx inRR# So the critical points in #kpi# are local minimum.

Elijah Abram · Answer

undefined To find all local maximum and minimum points using the second derivative test for $y = \tan^2(x)$:

1. Find the first derivative of $y$ with respect to $x$, $y'$.
2. Find the critical points by solving $y' = 0$ or $y'$ undefined.
3. Compute the second derivative of $y$ with respect to $x$, $y''$.
4. Evaluate $y''$ at each critical point.
5. If $y''$ is positive at a critical point, the point corresponds to a local minimum. If $y''$ is negative at a critical point, the point corresponds to a local maximum. If $y''$ equals zero, the test is inconclusive.

Given $y = \tan^2(x)$:
1. $y' = 2\tan(x)\sec^2(x)$
2. Critical points occur where $2\tan(x)\sec^2(x) = 0$. These points are $x = k\pi$ where $k$ is an integer.
3. $y'' = 2\sec^4(x) + 4\tan^2(x)\sec^2(x)$
4. Evaluate $y''$ at each critical point.
5. For $x = k\pi$, $y'' = 2\sec^4(k\pi) + 4\tan^2(k\pi)\sec^2(k\pi)$.

Since $\tan(k\pi) = 0$ and $\sec(k\pi) = (-1)^k$, we have:

- If $k$ is even, then $\tan(k\pi) = 0$ and $\sec(k\pi) = 1$, so $y'' = 2$. This implies a local minimum at $x = k\pi$.
- If $k$ is odd, then $\tan(k\pi) = 0$ and $\sec(k\pi) = -1$, so $y'' = -2$. This implies a local maximum at $x = k\pi$.

Therefore, all local maximum points occur at $x = (2n+1)\pi$ for integer $n$, and all local minimum points occur at $x = 2n\pi$ for integer $n$.