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The local minimum value of the function \(f^{\prime}\) given by \(f(x)=3+|x|\), \(x \in R\) is.
Options:
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\( \square 1 \)
Given that, $f(x)=3+|x|$ We know that $|x|=\sqrt{x^{2}}$ So, $f(x)=3+\sqrt{x^{2}}$
So, $f^{\prime}(x)=\frac{2 x}{2 \sqrt{x^{2}}}=\frac{x}{|x|}$
Clearly when $x>0$,local minimum $=1$
When $x < 0$, local minimum $=-1$
So, option (3) and (4) both are true
So, $f^{\prime}(x)=\frac{2 x}{2 \sqrt{x^{2}}}=\frac{x}{|x|}$
Clearly when $x>0$,local minimum $=1$
When $x < 0$, local minimum $=-1$
So, option (3) and (4) both are true
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