Question

Examine the following functions for continuity :

(a) f(x) = x - 5

(b) $f\left(x\right)=\frac{1}{x-5},x\ne 5$

(c) $f\left(x\right)=\frac{x^2-25}{x+5},x\ne -5$

(d) $f\left(x\right)=|x-5|$.

Answer 1

f(x) = x - 5 is a polynomial functions, so f(x) is continuous for all values fo x.

$f\left(x\right)=\frac{1}{x-5}$ is a quotient functions of two polynomial functions, so f(x) is continuous for all values of x provided $x\ne 5$.

Note In any rational function $f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}$ and if p(x) and q(x) are polynomials and q(x) is zero at any value of x and p(x) = 0, then f(x) is not continuous at that point.

$f\left(x\right)=\frac{x^2-25}{x+5}=\frac{(x+5)(x-5)}{(x+5)}=x-5$

$\therefore f\left(x\right)=x-5$ is a polynomial function, so f(x) is continuous at all values of x.

Note In any rational function $f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}$ and if P(x) and q(x) are polynomials and q(x) is zero at any value of x and p(x) = 0, then f(x) is not continuous at that point.

$f\left(x\right)=|x-5|=\{\begin{array}{c}x-5,forx\ge 5\\ 5-x,forx<5\end{array}$

$Forx\to 5^{+},$ $\underset{x\to 5^{+}}{lim}$ = $\underset{x\to 5^{+}}{lim}$ (x-5)=5-5=0

$Forx\to 5^{-},$ $\underset{x\to 5^{-}}{lim}$ f(x) = $\underset{x\to 5^{-}}{lim}$ (5-x)=5-5=0

Also, f(5)=5-5=0

$\therefore$ LHL=RHL=f(x). Therefore, the function is continuous at x=5.

Note In any rational function $f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}$ and if p(x) and q(x) are polynomials and q(x) is zero at any value of x and p(x) = 0, then f(x) is not continuous at that point.