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Math  /  Algebra

QuestionA function f(x)f(x) is defined by f(x)={x2−1x−1,x<1a2−7,x=1∣1−x∣x−1+b,x>1f(x)=\left\{\begin{array}{ll} \frac{x^{2}-1}{x-1} & , x<1 \\ a^{2}-7 & , x=1 \\ \frac{|1-x|}{x-1}+b & , x>1 \end{array}\right.
Determine the values of the constants aa and bb if f(x)f(x) is continous at x=1x=1.

Studdy Solution

STEP 1

1. The function f(x) f(x) is piecewise defined with three parts.
2. For f(x) f(x) to be continuous at x=1 x = 1 , the left-hand limit, the value of the function at x=1 x = 1 , and the right-hand limit must all be equal.
3. We need to find the values of a a and b b such that these conditions are satisfied.

STEP 2

1. Simplify and find the left-hand limit as x→1− x \to 1^- .
2. Evaluate the function at x=1 x = 1 .
3. Simplify and find the right-hand limit as x→1+ x \to 1^+ .
4. Set the left-hand limit, function value at x=1 x = 1 , and right-hand limit equal and solve for a a and b b .

STEP 3

Simplify the expression for x<1 x < 1 :
f(x)=x2−1x−1 f(x) = \frac{x^2 - 1}{x - 1}
Recognize that x2−1 x^2 - 1 is a difference of squares:
x2−1=(x−1)(x+1) x^2 - 1 = (x - 1)(x + 1)
Thus, the expression simplifies to:
f(x)=x+1forx<1 f(x) = x + 1 \quad \text{for} \quad x < 1

STEP 4

Evaluate the function at x=1 x = 1 :
f(1)=a2−7 f(1) = a^2 - 7

STEP 5

Simplify the expression for x>1 x > 1 :
f(x)=∣1−x∣x−1+b f(x) = \frac{|1-x|}{x-1} + b
Recognize that for x>1 x > 1 , ∣1−x∣=−(x−1) |1-x| = -(x-1) , so:
f(x)=−(x−1)x−1+b=−1+b f(x) = \frac{-(x-1)}{x-1} + b = -1 + b

STEP 6

Set the left-hand limit, function value at x=1 x = 1 , and right-hand limit equal:
1. Left-hand limit as x→1− x \to 1^- : lim⁡x→1−f(x)=1+1=2 \lim_{x \to 1^-} f(x) = 1 + 1 = 2
2. Value at x=1 x = 1 : f(1)=a2−7 f(1) = a^2 - 7
3. Right-hand limit as x→1+ x \to 1^+ : lim⁡x→1+f(x)=−1+b \lim_{x \to 1^+} f(x) = -1 + b
Set these equal:
2=a2−7=−1+b 2 = a^2 - 7 = -1 + b
Solve the equations:
1. a2−7=2 a^2 - 7 = 2 a2=9 a^2 = 9 a=±3 a = \pm 3
2. −1+b=2 -1 + b = 2 b=3 b = 3
The values of the constants are a=±3 a = \pm 3 and b=3 b = 3 .

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