The function f(x) = x/2 + 2/x has a local minimum at (A) x = 2 (B) x = -2 (C) x = 0 (D) x = 1

Q: Wait, why do we reject the negative root here?

We actually don't reject it! We must consider both $x = 2$ and $x = -2$ as critical points. However, when we test them using the second derivative, we find that $x = -2$ corresponds to a local maximum, not a minimum. Always check the nature of your critical points!

Q: Can I use AM-GM inequality instead of calculus?

That is a brilliant observation! For $x > 0$, the AM-GM inequality states that $\frac{x}{2} + \frac{2}{x} \ge 2\sqrt{\frac{x}{2} \cdot \frac{2}{x}} = 2$. The equality holds when $\frac{x}{2} = \frac{2}{x}$, which gives $x^2 = 4$, or $x = 2$. This is a faster, more elegant way to solve it for positive $x$!

Q: What happens at $x = 0$?

At $x = 0$, the function is undefined because we cannot divide by zero. Graphically, this creates a vertical asymptote. The function approaches infinity as $x$ approaches zero from the right, and negative infinity from the left. It is not a local minimum.

Question

The function $f(x) = \frac{x}{2} + \frac{2}{x}$ has a local minimum at

Options:

(A) $x = 2$

(B) $x = -2$

(C) $x = 0$

(D) $x = 1$

Sigma Percentile · Accepted Answer

### Analyzing the Landscape Welcome, future engineers! Today, we are going to explore the function $f(x) = \frac{x}{2} + \frac{2}{x}$. Imagine you are standing on a landscape defined by this curve. As you move along the $x$-axis, the function values rise and fall, creating a unique terrain. Our mission is to find the lowest point, the local minimum, on this curve. This is not just about solving an equation; it is about understanding the geometry of change. ### The Toolkit: Derivatives as Tangent Slopes To find the 'valley' of our function, we need to know where the curve flattens out. In calculus, the first derivative $f'(x)$ tells us the slope of the tangent line at any point. When the curve reaches a local minimum or maximum, the tangent line becomes perfectly horizontal, meaning its slope is zero. To make our differentiation easier, we rewrite the function using negative exponents: $$f(x) = \frac{1}{2}x + 2x^{-1}$$ Now, applying the power rule term by term, we find the derivative: $$f'(x) = \frac{1}{2} - 2x^{-2}$$ This simplifies to: $$f'(x) = \frac{1}{2} - \frac{2}{x^2}$$ ### The Search for Critical Points Now, we set our derivative to zero to find the critical points: $$\frac{1}{2} - \frac{2}{x^2} = 0$$ By moving the negative term to the right, we get: $$\frac{1}{2} = \frac{2}{x^2}$$ Cross-multiplying gives us $x^2 = 4$. Here is where many students stumble! When we take the square root of $4$, we must consider both the positive and negative roots: $x = 2$ and $x = -2$. These are our critical points, the locations where the curve is flat. ### The Second Derivative Test: Concavity We have two candidates, but which one is the valley? We use the second derivative test to check the concavity. Differentiating $f'(x) = \frac{1}{2} - 2x^{-2}$ again, we get: $$f''(x) = 0 - 2(-2)x^{-3} = \frac{4}{x^3}$$ The second derivative tells us about the 'smile' or 'frown' of the curve. If $f''(x) > 0$, the curve is concave up (a smile), indicating a local minimum. If $f''(x) < 0$, it is concave down (a frown), indicating a local maximum. Testing $x = 2$, we find: $$f''(2) = \frac{4}{2^3} = \frac{4}{8} = \frac{1}{2}$$ Since $\frac{1}{2} > 0$, the curve is concave up at $x = 2$, confirming it is a local minimum. Conversely, testing $x = -2$ yields $f''(-2) = -\frac{1}{2} < 0$, confirming it is a local maximum. Thus, the **local minimum is at $x = 2$**. Keep exploring, keep questioning, and let the math guide you!

Animated Solution for Mathematics - Differentiation: The function $f (x) = \frac{x}{2} + \frac{2}{x}$ has a local minimum at

Select Answer:

Visualized Solution

Visualizing $f (x) = \frac{x}{2} + \frac{2}{x}$

The First Derivative Tool

Preparing for Differentiation

Differentiating $f (x)$

Setting Derivative to Zero

Solving for $x$

The Second Derivative Test

Calculating $f^{''} (x)$

Testing $x = 2$

Conclusion

The Sigma Insight: Maxima and Minima

Analyzing the Landscape

The Toolkit

The Search for Critical Points

The Second Derivative Test

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function f(x)=2x​+x2​ has a local minimum at

Select Answer:

Visualized Solution

Visualizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)=2x​+x2​

The First Derivative Tool

Preparing for Differentiation

Differentiating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Setting Derivative to Zero

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

The Second Derivative Test

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′′(x)

Testing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2

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The Toolkit

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Animated Solution for Mathematics - Differentiation: The function $f (x) = \frac{x}{2} + \frac{2}{x}$ has a local minimum at

Visualizing $f (x) = \frac{x}{2} + \frac{2}{x}$

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Solving for $x$

Calculating $f^{''} (x)$

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