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Find the minimum or the minimum values of the function y=x+1/x for x>0.

Jayant Kumar , 11 Years ago
Grade 11
anser 1 Answers
Askiitians Tutor Team

Last Activity: 6 Days ago

To find the minimum value of the function \( y = x + \frac{1}{x} \) for \( x > 0 \), we can use calculus, specifically the concept of derivatives. This function is interesting because it combines a linear term and a reciprocal term, which can behave quite differently as \( x \) changes. Let's break it down step by step.

Step 1: Differentiate the Function

First, we need to find the derivative of the function with respect to \( x \). The derivative will help us identify the critical points where the function could have a minimum or maximum value.

The derivative of \( y \) is calculated as follows:

  • For the term \( x \), the derivative is \( 1 \).
  • For the term \( \frac{1}{x} \), we use the power rule: \( \frac{d}{dx}(x^{-1}) = -x^{-2} = -\frac{1}{x^2} \).

Putting it together, we have:

y' = 1 - \frac{1}{x^2}

Step 2: Set the Derivative to Zero

To find the critical points, we set the derivative equal to zero:

1 - \frac{1}{x^2} = 0

Solving for \( x \), we rearrange the equation:

\frac{1}{x^2} = 1

This implies:

x^2 = 1

Taking the square root gives us:

x = 1

Since we are only considering \( x > 0 \), we discard the negative root.

Step 3: Determine if it is a Minimum

Next, we need to confirm whether this critical point is a minimum. We can do this by examining the second derivative:

Taking the derivative of \( y' \):

y'' = \frac{2}{x^3}

Since \( x > 0 \), \( y'' \) is always positive. This indicates that the function is concave up at \( x = 1 \), confirming that we have a local minimum.

Step 4: Calculate the Minimum Value

Now that we know \( x = 1 \) is a minimum, we can find the minimum value of the function by substituting \( x = 1 \) back into the original function:

y(1) = 1 + \frac{1}{1} = 2

Final Thoughts

Thus, the minimum value of the function \( y = x + \frac{1}{x} \) for \( x > 0 \) is 2, occurring at \( x = 1 \). This result illustrates how the balance between the linear growth of \( x \) and the decreasing nature of \( \frac{1}{x} \) leads to a minimum point, showcasing the interplay of different mathematical behaviors in a single function.

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