**Application of
Extremum**

**Application of Extremum: **

1. Examine the point of intersection of y = f(x) with x – axis and y – axis

2. Examine whether denominator has a root or not. If no, then graph is continuous and f is non monotonic,

**Examples: **

f(x) = x/ (x² + x + 1)

f(x) = (x² + x- 2)/ (x² + x + 1)

If denominator has roots, then f(x) is discontinuous. Such
function can be monotonic/ non monotonic, **Example**

f(x) = (x² – x)/ (x² – 3x – 4)

3. If numerator and denominator have a common factor (say x = a) then y = f(x) removable discontinuity and will at x = a

**Example:**

f(x) = (x² – x)/ (x² – 3x + 2)

= (x(x-1))/ (x² – 3x + 2)

Functions of type linear/linear represent rectangular hyperbola excluding the point of discontinuity and will always be monotonic.

4. Compute dy/dx and find intervals where f(x) is increasing

5. At the point of discontinuity (say x = 1), check the limiting values \(\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f(x)\) and \(\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f(x)\) to find whether f approaches \(\infty \) (or) \(-\infty \)

**Example: **Draw the graph of f(x) = (x² – x + 1)/(x² +
x +1)

**Solution:**

Given that

f(x) = (x² – x + 1)/ (x² + x +1)

The given function is continuous for all x ϵ R

f’(x) = 2(x² – 1) / (x² + x +1) ²

f’(x)= 0

→ x = ± 1

x = 1 is the point of minima and x = -1 is the point of maxima

also, f(1) = 1/3,

f(-1) = 3 and

f(0) = 1

\(\underset{x\to \pm \infty }{\mathop{\lim }}\,\frac{{{x}^{2}}-x+1}{{{x}^{2}}+x+1}=1\)