Question

The local minimum value of the function $f(x)=x^{2}-x$, $x\in\mathbb{R}$, is

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{-1}{4}$
• D. $\frac{3}{4}$
• E. $\frac{-1}{2}$

Hint:

Algebra Tip: Because $y = x^2 - x$ is a standard upward-opening parabola, its minimum value is just the y-coordinate of its vertex. Using the vertex formula $x = -b/(2a)$, we instantly get $x = -(-1)/2 = 1/2$.

Answer 1

The Correct Option is C

Concept:
Local extrema occur at critical points where the first derivative is zero ($f^{\prime}(x) = 0$). To confirm whether a critical point is a minimum, check the second derivative; if $f^{\prime\prime}(x) > 0$, it is a local minimum. Finally, plug the critical x-value back into the original function to find the minimum value.

Step 1: Find the first derivative.
Differentiate the function $f(x) = x^2 - x$ with respect to $x$: $$f^{\prime}(x) = 2x - 1$$

Step 2: Find the critical point(s).
Set the first derivative equal to zero and solve for $x$: $$2x - 1 = 0$$ $$2x = 1 \implies x = \frac{1}{2}$$

Step 3: Verify it is a local minimum.
Find the second derivative to check concavity: $$f^{\prime\prime}(x) = 2$$ Since $2 > 0$ for all $x$, the curve is concave up everywhere, confirming that $x = \frac{1}{2}$ is indeed a local minimum.

Step 4: Substitute into the original function.
To find the actual minimum value, plug $x = \frac{1}{2}$ back into $f(x)$: $$f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)$$

Step 5: Calculate the final numerical value.
Evaluate the squares and subtract: $$f\left(\frac{1}{2}\right) = \frac{1}{4} - \frac{1}{2}$$ Get a common denominator: $$f\left(\frac{1}{2}\right) = \frac{1}{4} - \frac{2}{4} = \frac{-1}{4}$$ Hence the correct answer is (C) $\frac{-1{4}$}.