If the real number \( x \) when added to its inverse gives the minimum value of the sum, then the value of \( x \) is equal to
Finding the Minimum Value of \( x + \frac{1}{x} \)
We need to minimize the function:
\[
f(x) = x + \frac{1}{x}
\]
Step 1: Find the First Derivative
To find the critical points, differentiate \( f(x) \):
\[
f'(x) = 1 - \frac{1}{x^2}
\]
Setting \( f'(x) = 0 \):
\[
1 - \frac{1}{x^2} = 0
\]
\[
\frac{1}{x^2} = 1
\]
\[
x^2 = 1
\]
\[
x = \pm 1
\]
Step 2: Second Derivative Test
Differentiate \( f'(x) \) again:
\[
f''(x) = \frac{2}{x^3}
\]
Evaluate at \( x = 1 \):
\[
f''(1) = \frac{2}{1^3} = 2 > 0
\]
Since \( f''(1) > 0 \), \( x = 1 \) is a local minimum.
Final Answer:
\(
{1}
\)
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