Question

Find the critical points of the function $y = \tan^{-1}(\sec x)$ where $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. Show your steps.

Answer 1

x=0

Answer 2

To find the critical points of a function $y = f(x)$, we need to find the values of $x$ in the domain of the function where either $f'(x) = 0$ or $f'(x)$ is undefined.

1. Determine the Domain of the Function: The given function is $y = \tan^{-1}(\sec x)$. The domain of $\tan^{-1}(u)$ is all real numbers, $u \in (-\infty, \infty)$. The function $\sec x = \frac{1}{\cos x}$ is defined when $\cos x \neq 0$. In the interval $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, $\cos x = 0$ at $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$. Therefore, the function $y = \tan^{-1}(\sec x)$ is defined for $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$. We will look for critical points within this open interval.

2. Calculate the First Derivative ($y'$): We use the chain rule for differentiation. Let $u = \sec x$. Then $y = \tan^{-1}(u)$. The derivative of $\tan^{-1}(u)$ with respect to $u$ is $\frac{1}{1+u^2}$. The derivative of $\sec x$ with respect to $x$ is $\sec x \tan x$. Applying the chain rule, $y' = \frac{dy}{du} \cdot \frac{du}{dx}$: $y' = \frac{1}{1+(\sec x)^2} \cdot (\sec x \tan x)$ $y' = \frac{\sec x \tan x}{1+\sec^2 x}$

3. Find Points where $y' = 0$: Set the derivative equal to zero: $\frac{\sec x \tan x}{1+\sec^2 x} = 0$ The denominator $1+\sec^2 x$ is always positive because $\sec^2 x \ge 1$ for $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$. Thus, $1+\sec^2 x \ge 2$, and it is never zero. Therefore, for $y'$ to be zero, the numerator must be zero: $\sec x \tan x = 0$ This implies either $\sec x = 0$ or $\tan x = 0$.

• $\sec x = 0 \implies \frac{1}{\cos x} = 0$, which is impossible for any real value of $x$.
• $\tan x = 0$. In the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, $\tan x = 0$ only when $x = 0$.

So, $x=0$ is a critical point.

4. Find Points where $y'$ is Undefined: The derivative $y' = \frac{\sec x \tan x}{1+\sec^2 x}$ would be undefined if:

• The denominator $1+\sec^2 x = 0$. As established, this is never true.
• $\sec x$ or $\tan x$ are undefined. Both $\sec x$ and $\tan x$ are undefined at $x = \pm \frac{\pi}{2}$. However, these points are outside the domain of the original function $y = \tan^{-1}(\sec x)$. Within the domain $(-\frac{\pi}{2}, \frac{\pi}{2})$, $\sec x$ and $\tan x$ are always defined, and thus $y'$ is always defined.

Therefore, there are no critical points where the derivative is undefined.

Conclusion: The only critical point of the function $y = \tan^{-1}(\sec x)$ in the given interval is $x=0$.