# Prove that the function f(x) = 5x − 3 is continuous at x = 0, x = −3 and at x = 5

**Solution:**

A function is said to be continuous when the graph of the function is a single unbroken curve.

The given function is f(x) = 5x − 3

At x = 0,

f(0) = 5(0) − 3 = −3

lim_{x→0 }f(x) = lim_{x→0 }(5x − 3)

= 5(0) −3 = −3

⇒ lim_{x→0 }f(x) = f(0)

Therefore,

f is continous at x = 0

At x = −3,

f(−3) = 5(−3) − 3 = −18

lim_{x→−3 }f(x) = lim_{x→−3 }(5x − 3)

= 5(−3) − 3 = −18

⇒ lim_{x→−3 }f(x) = f(−3)

Therefore,

f is continous at x = −3.

At x = 5,

f(5) = 5(5) − 3 = 22

lim_{x→5} f(x) = lim_{x→5} (5x − 3) = 5(5) − 3 = 22

⇒ lim_{x→5} f(x) = f(5)

Therefore, f is continous at x = 5

NCERT Solutions Class 12 Maths - Chapter 5 Exercise 5.1 Question 1

## Prove that the function f(x) = 5x − 3 is continuous at x = 0, x = −3 and at x = 5

**Summary:**

The function f(x) = 5x − 3 is continuous at x = 0, x = −3 and at x = 5

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