Question

Prove that the functions f(x) =5x-3 is continuous at x = 0, at x = -3 and at x=5.

Answer 1

Here, f(x) = 5x-3

At x=o, ${lim}_{x\to 0}$ f(x) = ${lim}_{x\to 0}$ (5x-3) = 5 x 0 - 3 = 0 - 3 = - 3

f(0)=5 x 0 - 3 = - 3

${lim}_{x\to 0}$ f(x) = f(0), Thus, f(x) is continuous at x=0.

At x = -3, ${lim}_{x\to -3}$ f(x) = ${lim}_{x\to -3}$ (5x-3) = 5 $\times (-3)-3$ = - 15 -3 = -18

and f(-3) = 5x-3-3=-18

${lim}_{x\to -3}$ f(x) = f(-3), Thus, f(x) is continuous at x=-3.

At x =5, ${lim}_{x\to 5}$ f(x) = ${lim}_{x\to 5}$ (5x-3) = 5$\times$ 5 - 3 = 25 -3 = 22

and f(5) = 5$\times$ 5 - 3 = 25 -3 = 22

${lim}_{x\to 5}$ f(x) = f(5). Thus, f(x) is continuous at x=5,