A function f(x) is defined by f(x) = \begin{cases} \frac{2x^2 - 5x - 3}{x - 3}, & x \neq 3 \\ k + 2, & x = 3 \end{cases} If f(x) is continuous at x = 3, what is the value of the constant k?

A function f(x) is defined by f(x) = \begin{cases} \frac{2x^2 - 5x - 3}{x - 3}, & x \neq 3… — JAMB UTME

A function

f(x)

is defined by

f(x) = \begin{cases} \frac{2x^2 - 5x - 3}{x - 3}, & x \neq 3 \\ k + 2, & x = 3 \end{cases}

f(x)

is continuous at

x = 3

, what is the value of the constant

k

The constant value is

k = 5

For the function to be continuous at

x = 3

, the limit as

x \to 3

must equal the value of the function at

x = 3

, which is

f(3) = k + 2

. Factoring the numerator gives

2x^2 - 5x - 3 = (2x + 1)(x - 3)

. Canceling the common factor

(x - 3)

leaves

\lim_{x \to 3}(2x + 1) = 7

. Setting

k + 2 = 7

yields

k = 5

Evaluate the limit of

f(x)

x

approaches

3

\lim_{x \to 3} \frac{2x^2 - 5x - 3}{x - 3} = \lim_{x \to 3} \frac{(2x + 1)(x - 3)}{x - 3} = \lim_{x \to 3} (2x + 1) = 2(3) + 1 = 7

Direct substitution yields the indeterminate form

\frac{0}{0}

, so the numerator must be factored to cancel the common term

(x - 3)

Apply the definition of continuity at

x = 3

f(3) = \lim_{x \to 3} f(x) \implies k + 2 = 7

For a function to be continuous at a point

x = c

, the function value

f(c)

must equal the limit of

f(x)

x \to c

Solve for the constant

k

k = 7 - 2 = 5

Subtract

2

from both sides of the equation.

Continuity of a Piecewise Function at a Point