Calculus is a very powerful mathematical tool. While all GMAT Quant questions can be solved without a knowledge of calculus, having the know how of this particular technique can make some questions ridiculously easy.

Lets start with an example :

**What is the minimum value of x ^{2} – 2x + 4?**

**A) 2**

**B) 3**

**C) 1**

**D) 1.5**

**E) 4**

A traditional approach would have us either factor the quadratic equation or draw a rough graph to figure out where the curve hits its bottom. While this works, it is time consuming.

Enter calculus!

First some basics:

A derivative of an equation is defined as a slope of that equation.

To find an inflexion point (i.e., minima or maxima) of an equation, we equate the slope of the equation to zero and solve for the variable.

In other words, for an equation f(x) and its slope f’(x), to find the inflexion point, we have to solve for f’(x) = 0

Now, to start of this, how do we even know what f’(x) is. Well, there are a few standard definitions.

**f(x) = x**^{n}then f’(x) = nx^{n-1}**f(x) = constant, then f’(x) = 0 (think about it, a constant would be a horizontal line on the Cartesian plane, hence slope = 0)****if f(x) = g(x) + h(x), then f’(x) = g’(x) + h’(x) (derivatives are additive and subtractive)**

There are multiple such definitions but the above three are more or less sufficient for the GMAT

Coming to the question.

f(x) = x^{2} – 2x + 4

f’(x) = 2x – 2

solving for f’(x) = 0, 2x -2 = 0 and x = 1

This is the value of x where f(x) either reaches a minima or a maxima. Let’s plug it back into the equation.

f(1) = 1^{2} – 2×1 + 4 = 3

Now intuitively we know this value (3) cannot represent a maximum, as the value of the function can take values greater than 3. Just plug any large number (e.g., 5) into the function and you will notice this behavior. Hence, 3 represent the minima and is the answer we are after.

**Answer (B). **Impressed, aren’t we 🙂

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