If the function f(x) = begin{cases} tan^{-1}x | vedclass

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(C) For a function $f(x)$ to have a local minimum at $x = a$,the value of the function at $x = a$ must be less than or equal to the values of the func

Vedclass · Accepted Answer

$\left( -\infty, \frac{\pi}{4} \right]$

Vedclass · Answer

(C) For a function $f(x)$ to have a local minimum at $x = a$,the value of the function at $x = a$ must be less than or equal to the values of the function in the immediate neighborhood of $a$.Specifically,for $x = 1$,we require $f(1) \le f(1 - h)$ and $f(1) \le f(1 + h)$ for a small positive $h$.First,calculate $f(1)$ using the definition for $x \ge 1$: $f(1) = \sec^{-1}(1) + \lambda = 0 + \lambda = \lambda$.Next,consider the limit as $x 	o 1^-$: $\lim_{x 	o 1^-} f(x) = 	an^{-1}(1) = \frac{\pi}{4}$.For $x = 1$ to be a local minimum,the value at $x = 1$ must be less than or equal to the values just to the left of $1$. Thus,we require $\lambda \le \frac{\pi}{4}$.Also,for $x > 1$,$f(x) = \sec^{-1}(x) + \lambda$. Since $\sec^{-1}(x)$ is an increasing function for $x \ge 1$,$f(x) > f(1)$ for all $x > 1$ regardless of $\lambda$.Therefore,the condition for a local minimum at $x = 1$ is simply $\lambda \le \frac{\pi}{4}$,which corresponds to the interval $\left( -\infty, \frac{\pi}{4} ight]$.

If the function $f(x) = \begin{cases} \tan^{-1}x; & x < 1 \\ \sec^{-1}x + \lambda; & x \ge 1 \end{cases}$ has a local minimum at $x = 1$,then the range of $\lambda$ is:

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