A piecewise function f(x) is defined by f(x) = \begin{cases} \frac{2x^2 - 5x - 3}{x - 3}, & \text{for } x \neq 3 \ a^2 - 2, & \text{for } x = 3 \end{cases} If f(x) is continuous at x = 3 and a > 0, what is the numerical value of a?

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

A piecewise function

f(x)

is defined by

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

f(x)

is continuous at

x = 3

and

a > 0

, what is the numerical value of

a

Cevap: 3

Cevap

The numerical value of a is 3.

For

f(x)

to be continuous at

x = 3

, the defined value

f(3) = a^2 - 2

must equal

\lim_{x \to 3} f(x)

. Factoring the numerator gives

\frac{(2x + 1)(x - 3)}{x - 3} = 2x + 1

for

x \neq 3

. Taking the limit as

x \to 3

yields

2(3) + 1 = 7

. Setting

a^2 - 2 = 7

leads to

a^2 = 9

, which gives

a = 3

under the constraint

a > 0

Adım Adım Çözüm

Evaluate the limit of

f(x)

x \to 3

Factor the numerator

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

. For

x \neq 3

f(x) = 2x + 1

. Thus,

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

Direct substitution gives an indeterminate form

\frac{0}{0}

, so canceling the common factor

(x - 3)

allows direct evaluation of the limit.

Apply the definition of continuity at

x = 3

f(3) = a^2 - 2 = 7

For a function to be continuous at a point

c

, the function value

f(c)

must equal the limit

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

Solve for the parameter

a

a^2 = 9 \implies a = 3

(since

a > 0

Solving

a^2 = 9

gives solutions

3

and

-3

. The condition

a > 0

specifies the positive root.

Anahtar Kavram

Continuity of a Piecewise Function at a Point

Soru

Cevap

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Anahtar Kavram

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