Question

let c be a positive constant. the function ( f(x)=sqrt{|x|+c} ) has... (hint: consider ( f(x) ) as a piecewise function.) a local minimum located at a singular point. a local minimum located at a critical point. a local maximum located at a critical point. no local extrema a local maximum located at a singular point.

Explanation:

Step1: Write the piece - wise function

When \(x\geq0\), \(f(x)=\sqrt{x + c}\); when \(x<0\), \(f(x)=\sqrt{-x + c}\).

Step2: Find the derivative for each piece

For \(x>0\), using the chain rule \(y = \sqrt{u}\), \(u=x + c\), \(y^\prime=\frac{1}{2\sqrt{x + c}}\).
For \(x<0\), using the chain rule \(y=\sqrt{u}\), \(u=-x + c\), \(y^\prime=\frac{-1}{2\sqrt{-x + c}}\).
At \(x = 0\), the left - hand derivative \(\lim_{x
ightarrow0^{-}}\frac{f(x)-f(0)}{x-0}=\lim_{x
ightarrow0^{-}}\frac{\sqrt{-x + c}-\sqrt{c}}{x}\). Multiply numerator and denominator by \(\sqrt{-x + c}+\sqrt{c}\), we get \(\lim_{x
ightarrow0^{-}}\frac{-x}{x(\sqrt{-x + c}+\sqrt{c})}=-\frac{1}{2\sqrt{c}}\).
The right - hand derivative \(\lim_{x
ightarrow0^{+}}\frac{f(x)-f(0)}{x - 0}=\lim_{x
ightarrow0^{+}}\frac{\sqrt{x + c}-\sqrt{c}}{x}\). Multiply numerator and denominator by \(\sqrt{x + c}+\sqrt{c}\), we get \(\lim_{x
ightarrow0^{+}}\frac{x}{x(\sqrt{x + c}+\sqrt{c})}=\frac{1}{2\sqrt{c}}\).
Since the left - hand and right - hand derivatives at \(x = 0\) are not equal, \(x = 0\) is a singular point.

Step3: Analyze the function values around \(x = 0\)

For \(x>0\), \(f(x)=\sqrt{x + c}>\sqrt{c}\) (because \(x>0\)). For \(x<0\), \(f(x)=\sqrt{-x + c}>\sqrt{c}\) (because \(x<0\) implies \(-x>0\)). And \(f(0)=\sqrt{c}\).

Answer:

a local minimum located at a singular point.