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Question: The function \( f(x) = \dfrac{{{x^2} - 9}}{{x - 3}} \) is not defined at \( x = 3 \) . What should b...

The function f(x)=x2−9x−3f(x) = \dfrac{{{x^2} - 9}}{{x - 3}} is not defined at x=3x = 3 . What should be assigned to f(3)f\left( 3 \right) for continuity of f(x)f(x) at x=3x = 3 ?

Explanation

Solution

Continuity at a point can be determined by substituting the point in the function. If this gives a definite value then the function is said to be continuous at that point and if it is not defined then it is not continuous.
At x=3x = 3 the function, f(x)=x2−9x−3f(x) = \dfrac{{{x^2} - 9}}{{x - 3}} is not defined, still we can find the limit at which function is continuous.
Find f(3)=lim⁡x→3x2−9x−3f(3) = \mathop {\lim }\limits_{x \to 3} \dfrac{{{x^2} - 9}}{{x - 3}} by factorization of the numerator of f(x)f(x) .
Use the factorization formula, a2−b2=(a−b)(a+b){a^2} - {b^2} = (a - b)(a + b) .
Next, substitute x=3x = 3 for continuity of f(x)f(x) .

Complete step-by-step answer:
Consider the function, f(x)=x2−9x−3f(x) = \dfrac{{{x^2} - 9}}{{x - 3}} .
It is given that the function is not continuous at x=3x = 3 as the denominator becomes zero when we put x=3x = 3 .
We have to find the value f(3)f\left( 3 \right) for continuity of f(x)=x2−9x−3f(x) = \dfrac{{{x^2} - 9}}{{x - 3}} .
To be the function f(x)f(x) continuous, find the limit,
f(3)=lim⁡x→3x2−9x−3f(3) = \mathop {\lim }\limits_{x \to 3} \dfrac{{{x^2} - 9}}{{x - 3}}
Write 99 as 32{3^2} .
f(3)=lim⁡x→3x2−32x−3……(1)f(3) = \mathop {\lim }\limits_{x \to 3} \dfrac{{{x^2} - {3^2}}}{{x - 3}} \ldots \ldots (1)
Apply the factorization formula, a2−b2=(a−b)(a+b){a^2} - {b^2} = (a - b)(a + b) to the numerator of the right-hand side.
Substitute a=xa = x and b=3b = 3 into the factorization formula and a2−b2=(a−b)(a+b){a^2} - {b^2} = (a - b)(a + b) ,
x2−32=(x−3)(x+3)……(2){x^2} - {3^2} = (x - 3)(x + 3) \ldots \ldots (2)
From equation (1)(1) and (2)(2) .
f(3)=lim⁡x→3(x−3)(x+3)x−3f(3) = \mathop {\lim }\limits_{x \to 3} \dfrac{{(x - 3)(x + 3)}}{{x - 3}}
Cancel out the x−3x - 3 numerator of the right-hand side.
f(3)=lim⁡x→3x+3f(3) = \mathop {\lim }\limits_{x \to 3} x + 3
Apply the limit as xx tends to 33 ,
f(3)=3+3f(3) = 3 + 3
f(3)=6f(3) = 6

Final Answer: The value assigned to f(3)f\left( 3 \right) for continuity of f(x)f(x) at x=3x = 3 is 66 .

Note:
It is important to remember the definition of continuity
A function f(x)f(x) is said to be continuous at a point x=ax = a if lim⁡x→af(x)=f(a)\mathop {\lim }\limits_{x \to a} f(x) = f(a) and
If f(x)f(x) is continuous at x=ax = a then,
Limit at the point: lim⁡x→af(x)=f(a)\mathop {\lim }\limits_{x \to a} f(x) = f(a)
Left-hand limit: lim⁡x→a−f(x)=f(a)\mathop {\lim }\limits_{x \to {a^ - }} f(x) = f(a)
Right-hand limit: lim⁡x→a+f(x)=f(a)\mathop {\lim }\limits_{x \to {a^ + }} f(x) = f(a)