KCET · Maths · Functions
The local minimum value of the function \(f^{\prime}\) given by \(f(x)=3+|x|\), \(x \in R\) is.
- A \( \square 3 \)
- B \( \square 0 \)
- C \( \square -3 \)
- D \( \square 1 \)
Answer & Solution
Correct Answer
(D) \( \square 1 \)
Step-by-step Solution
Detailed explanation
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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