# Sketch the curve $$y = \frac{1}{x – 3}$$

Solution

Given that

$$y = \frac{1}{x – 3}$$

A. Domain

Clearly, domain is the set of all real number except 3.

i.e. domain is (-∞, 3) ∪ (3, ∞)

B. Intercept

For x-intercept: Put y = 0 we get x = ∞

For y-intercept: Put x = 0  we get $$\frac{-1}{3}$$

Thus curve meet the x-axis and y-axis at origin

C. Symmetry

Since f(-x) -f(x), so it is not symmetrical about origin.

D. Asymptotes

$$\lim_{x \to \pm ∞} f(x) = \lim_{x \to \pm ∞} \frac{1}{x-3} = ∞$$

Thus the curve doesn’t have horizontal assymptotes

$$\lim_{x \to 3} \frac{1}{x-3} = ∞$$

So, x = 3 is the vertical assymptotes.

E. Extreme

$$f^{‘}(x) = \frac{d}{dx} (x-3)^{-1}$$

$$f^{‘}(x) = -1 (x-3)^{-1-1}$$

$$f^{‘}(x) = \frac{1}{(x-3)^{2}}$$

$$f^{‘}(x) = -\frac{1}{(x-3)^2}$$

For critical points $$f^{‘}(x) = 0$$ and $$f^{‘}(x) = ∞$$

We get x = 3

 Interval (-∞, 3) (3,∞) Sign of f'(x) -ve +ve Nature of f(x) Decreasing Increasing

F. Concavity

For concavity and point of inflection

$$f^{”}(x) = 2 \times \frac{1}{(x-3)^3}$$

For point of inflection $$f^{‘}(x) = 0$$ and $$f^{‘}(x) = ∞$$

We get x = 3

 Interval (-∞, 3) (3,∞) Sign of $$f^{”}(x)$$ -ve -ve Nature of f(x) Concave Down Concave Down

G. Summary

 Interval (-∞, 3) (3,∞) Sign of f(x) Decreasing Increasing f(x) Concave Down Concave Down

Using this table with information we can sketch the graph as follows: