## JEE advanced important topics : derivative and growth of functions

JEE advanced important topics : The derivative of a function at a point indicates the rhythm of change of the function in said point in space. An immediate consequence is that the derivative will allow us to know if a function is increasing or decreasing in a determinate point, with only calculating the result of the derivative and looking if it’s greater or smaller than zero. If the derivative is positive (greater than zero), the variation of the function is positive, which means it’s growing in the point that we considered for the calculation. On the contrary, if the derivative is negative (smaller than zero), the variation in the function is negative, which means the function is decreasing in the considered point.

If we know the derivative function, the problem in determining the intervals in which the function increases and decreases is reduced to the study of the sign of it’s derivative function and nothing more. The intervals in which the derivative function is positive correspond to the intervals in which the primitive function is increasing. The intervals in which the derivative function is negative correspond to the intervals in which the primitive function is decreasing.

When the derivative in a point is zero, the tangent in the function in said point is horizontal. But the tangent can be horizontal because of several reasons, that is why interpreting the null derivative tends to be more complex than when it is positive or negative. So, it may happen that you find a relative maximum or a relative minimum, or it can be an inflection point in the horizontal tangent or it simply is a point in which the function is constant.

The cutting points in the derivative function with the OX axis, meaning, the points where the derivative function is equal to zero tell us where the primitive function has points with horizontal tangent and also help to determine the increasing and decreasing intervals of the function.

If we derive the derivative function we obtain the second derivative function. This will give us information about the variation of the last function, meaning, the first derivative function. Following the same reasoning we have made so far, the second derivative will allow us to know whether or not the first derivative is decreasing or increasing, which in turn allows to know the type of curvature in the primitive function, like in which intervals it’s convex or concave. In consequence, the second derivative will allow us to determine the intervals in which the function is concave or convex, and also the inflection points, which are the points in which the type of curvature of the function changes.

## Application: Increase and decrease

If f is a derivable function in a, which is one point in the dominion such as $f'(a) \neq 0$.

• f is strictly increasing in a if and only if f’(a)>0.
• f is strictly decreasing in a if and only if f’(a)<0.

### Example 1.

Study the increase and decrease of the function $f(x)=\left | x^{2}-1 \right |$

##### Resolution:

The function f can be express as:

$f(x) = \left\{ \begin{array}{ll} x^{2}-1 & x\leq -1 \\ 1-x^{2} & -1\leq x\leq 1 \\ x^{2}-1 & 1\leq x \\ \end{array} \right.$

The function is continuous for every real number and is not derivable in -1 and 1.

$f'(x) = \left\{ \begin{array}{ll} 2x & x\leq-1 \\ -2x & -1\leq x\leq 1 \\ 2x & 1\leq x \\ \end{array} \right.$

By studying the derivative, we can see that the points in which the derivative function can change from positive and negative (and so, the primitive function can go from increasing to decreasing) are – 1 , 0 and 1.

Note: The point 0 is calculated from the fact that only when x=0 then f’(x)=0, which means f’(x) in never zero for the interval ( – ∞, – 1) nor in the interval (1, ∞).

• In the interval (-∞, -1), you obtain that f’(x)<0, which means that in that interval the function decreases.
• In the interval (1, ∞), you obtain that f’(x)>0, which means that in that interval the function increases.

f’(x) is equal to zero in the interval (-1, 1), in the point 0, that is why we need to study both intervals, (-1, 0) and (0,1), separately.

• In the interval (-1, 0), you obtain that f’(x) = -2x >0, which means that in that interval the function increases.
• In the interval (0, 1), you obtain that f’(x)=-2x<0, which means that in that interval the function decreases.

To put it short, f increases in the intervals (-1,0) and (1, ∞) but f decreases in the intervals (-∞,-1) and (0,1). In the points -1, 0 and 1 the function remains constant, so it doesn’t increase nor it decreases.

### Example 2.

Find out in which intervals the function $f(x)=\frac{x^{2}-4x+1}{2}$ increases or decreases.

Resolution.

First, we need to calculate f’(x), f’(x)=(2x-4)/2 = x-2

As f’(x) > 0 ↔ x – 2>0, then x>2, so f is increasing for x>2.

As f’(x) < 0 ↔ x – 2<0, then x<2, so f is decreasing for x<2.

We can easily observe this in the graph of the function f shown here:

### Example 3.

Find out in which intervals the function $f(x)= x^{2}+\frac{1}{x^{2}}$ increases or decreases with x is not equal to zero.

Resolution.

$f'(x)= 2x-2\frac{1}{x^{3}}$, which can be written as: $f'(x)= \frac{2(x-1)(x+1)(x^{2}+1)}{x^{3}}$

As $2(x^{2}+1)$ is positive for all real numbers,

then: $latex f'(x)=\frac{(x-1)(x+1)}{x^{^{3}}} > 0 \Rightarrow x\in (-1,0)\cup (1,\infty)$
and
$f'(x)=\frac{(x-1)(x+1)}{x^{^{3}}} < 0 \Rightarrow x\in (-\infty,-1)\cup (0,1)$

Then: f’(x) > 0 if x ϵ [-1,0] U [1,+ ∞), which means that the function f increases in the interval [-1,0] U [1,+ ∞).

Also, f’(x)<0 if (-∞,-1] U [0,1] which means the function f decreases in the interval (-∞,-1] U [0,1].

The graphical representation of this function is shown in this diagram:

### Example 4.

Find out in which intervals the function f(x)=(x + 1)/(x – 1) increases or decreases with $x\neq 0$.

Resolution.

The derivative of f is $f'(x)=\frac{-2}{(x-1^{2})}$

As $(x-1)^{2}$ is bigger than zero for all the real numbers when $x\neq 1$, and also -2<0 then f’(x) < 0 for all the real numbers when $x\neq 1$. Here we have the graphical representation of this function as a visual aid: