Sketch the curve \(y = \frac{1}{x – 3}\)

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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:

User Loaded Image | CSIT Guide

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