Calculus

Application of Derivatives

Easy

core

The minimum value of $f(x) = |2x - 1|$ is

$-\infty$

$\frac{1}{2}$

✅ Correct Option: 2

$|2x - 1| \ge 0$ for all real x, with equality only when $2x - 1 = 0$ , i.e. $x = 1/2$ . So the minimum value is 0.

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In which of the following intervals, the function $f(x) = -x^2 - 2x + 15$ is decreasing?

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The total cost function is given by $c(x) = \frac{1}{3}x^3 - 5x^2 + 30x - 15$ and selling price per unit is Rs.6. The profit is maximum if the value of x is:

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The rate of change of the area of a circular disc with respect to its circumference when radius is 3 is :

The minimum value of f(x)=∣2x−1∣f(x) = |2x - 1|f(x)=∣2x−1∣ is

Solution

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The minimum value of $f(x) = |2x - 1|$ is