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Q.
The local minimum value of the function $f$ given by $f(x)=3+|x|, x \in R$ is
Application of Derivatives
Solution:
Note that the given function is not differentiable at $x=0$. So, second derivative test fails. Let us try first derivative test.
Note that 0 is a critical point of $f$. Now to the left of 0 , $f(x)=3-x$ and so $f^{\prime}(x)=-1<0$. Also to the right of $0_1 f(x)=3+x$ and so, $f^{\prime}(x)=1>0$. Therefore, by first derivative test, $x=0$ is a point of local minima of $f$ and local minimum value of $f$ is $f(0)=3$
