A function f(x) is defined by f(x) = \begin{cases} \frac{x^2 + x - 6}{x - 2}, & x \neq 2 \\ 2k - 1, & x = 2 \end{cases} If f(x) is continuous at x = 2, what is the value of the constant k?

A function f(x) is defined by f(x) = \begin{cases} \frac{x^2 + x - 6}{x - 2}, & x \neq 2 \… — JAMB UTME

A function

f(x)

is defined by

f(x) = \begin{cases} \frac{x^2 + x - 6}{x - 2}, & x \neq 2 \\ 2k - 1, & x = 2 \end{cases}

f(x)

is continuous at

x = 2

, what is the value of the constant

k

The value of the constant

k

3

By definition, a function

f(x)

is continuous at

x = c

\lim_{x \to c} f(x) = f(c)

. Factoring the numerator gives

x^2 + x - 6 = (x - 2)(x + 3)

. Canceling the common factor

(x - 2)

for

x \neq 2

, the limit as

x \to 2

2 + 3 = 5

. Equating

f(2) = 2k - 1

to 5 yields

2k - 1 = 5

, which solves to

k = 3

Evaluate the limit of

f(x)

x

approaches

2

\lim_{x \to 2} \frac{x^2 + x - 6}{x - 2} = \lim_{x \to 2} \frac{(x - 2)(x + 3)}{x - 2} = \lim_{x \to 2} (x + 3) = 5

Direct substitution gives the indeterminate form

\frac{0}{0}

, so factor the numerator to simplify.

Apply the definition of continuity at a point.

f(2) = \lim_{x \to 2} f(x) \implies 2k - 1 = 5

For

f(x)

to be continuous at

x = 2

, the value of the function at

x = 2

must equal its limit as

x \to 2

Solve the linear equation for

k

2k = 6 \implies k = 3

Add 1 to both sides and divide by 2.

Continuity of a Piecewise Function at a Point

Alternatively, use L'Hôpital's rule to evaluate the limit:

\lim_{x \to 2} \frac{\frac{d}{dx}(x^2+x-6)}{\frac{d}{dx}(x-2)} = \lim_{x \to 2} \frac{2x+1}{1} = 5

. Then set

2k - 1 = 5

to find

k = 3

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