Find the values of k so that the function f is continuous at the indicated point.
\(f(x)=\left\{\begin{matrix} kx+1, &if\, x\leq\pi \\ cos\,x,&if\,x>\pi \end{matrix}\right.\,at\,x=\pi\)
The given function is
\(f(x)=\left\{\begin{matrix} kx+1, &if\, x\leq\pi \\ cos\,x,&if\,x>\pi \end{matrix}\right.\)
The given function f is continuous at x=p, if f is defined at x=p and if the value of the f at x=p equals the limit of f at x=p.
It is evident that f is defined at x=p and f(π)=kπ+1
\(\lim_{x\rightarrow\pi^-}\) f(x)=\(\lim_{x\rightarrow\pi^+}\)f(x)=f(\(\pi\))
\(\Rightarrow\)\(\lim_{x\rightarrow\pi^-}\)(kx+1)=\(\lim_{x\rightarrow\pi^+}\)cosx=k\(\pi\)+1
\(\Rightarrow\)k\(\pi\)+1=cos\(\pi\)=k\(\pi\)+1
\(\Rightarrow\)k\(\pi\)+1=-1=k\(\pi\)+1
k=\(\frac{-2}{\pi}\)
Therefore, the required value of k is \(\frac{-2}{\pi}\).
The function \( f(x) \) is defined as follows: \[ f(x) = \begin{cases} 2+x, & \text{if } x \geq 0 \\ 2-x, & \text{if } x \leq 0 \end{cases} \] Then function \( f(x) \) at \( x=0 \) is: