\(f\left( x \right) = \begin{cases} e^{bx} & \text{for } \, x \leq 3 \\ \frac{2}{3}x + b & \text{for } \, x > 3 \end{cases}\)
Let f be the function defined above. For what values of b is f continuous at x=3?
A) 0.394 only
B) 0.274 only
C) −4.500 and 0.394
D) −1.998 and 0.274
▶️Answer/Explanation
Answer: D
Explanation:
For continuity at x=3, the left-hand limit (x→3⁻) must equal the right-hand limit (x→3⁺) and both must equal f(3).
1. Left-hand limit (x≤3): \( \lim_{x→3^-} f(x) = e^{3b} \)
2. Right-hand limit (x>3): \( \lim_{x→3^+} f(x) = \frac{2}{3}(3) + b = 2 + b \)
3. Set them equal: \( e^{3b} = 2 + b \)
4. Solving this equation gives two solutions: b ≈ -1.998 and b ≈ 0.274
Therefore, the correct answer is D) −1.998 and 0.274.
Let \( g \) be the function defined by:
\( g(x) = \begin{cases} \frac{x^2 – 4}{4x + 8} & \text{for } x \neq -2 \\ k & \text{for } x = -2 \end{cases} \)
For what value of \( k \) is \( g \) continuous at \( x = -2 \)?
A) -2
B) -1
C) -12
D) 0
▶️ Answer/Explanation
\(f(x) = \begin{cases} 2c + c \sin\left(\frac{\pi}{2}x\right) & \text{for } x < 3 \\ 7 & \text{for } x = 3 \\ 2c + 5x & \text{for } x > 3 \end{cases}\)
Let f be the function defined above. For what value of c, if any, is f continuous at x=3?
A) −4
B) 7
C) −15
D) There is no such c.
▶️Answer/Explanation
Answer: D
Explanation:
For continuity at x=3, the following must hold:
1. Left-hand limit (x→3⁻): \( \lim_{x→3^-} f(x) = 2c + c \sin\left(\frac{3\pi}{2}\right) = 2c – c = c \)
2. Right-hand limit (x→3⁺): \( \lim_{x→3^+} f(x) = 2c + 15 \)
3. Function value: \( f(3) = 7 \)
For continuity, all three must be equal:
\( c = 2c + 15 = 7 \)
This leads to two equations:
1. \( c = 7 \)
2. \( 2c + 15 = 7 \) ⇒ \( c = -4 \)
Since c cannot be both 7 and -4 simultaneously, there is no value of c that makes f continuous at x=3.
Which of the following are point(s) of discontinuity of the function \(f(x)=\frac{3x+1}{2x^{3}-8x^{2}-64x}\)?
I. 0
II. 4
III. 8
(A) I only
(B) II only
(C) I and II only
(D) I and III only
▶️Answer/Explanation
Answer: D
Explanation:
1. Find where the denominator equals zero:
\(2x^{3}-8x^{2}-64x = 0\)
Factor: \(2x(x^{2}-4x-32) = 0\)
Solutions: \(x = 0\) or \(x = \frac{4 \pm \sqrt{16+128}}{2} = \frac{4 \pm 12}{2}\)
Thus: \(x = 0\), \(x = 8\), or \(x = -4\)
2. The function is discontinuous at these points (I: 0 and III: 8)
3. At x = 4 (II), the function is continuous since the denominator is not zero
4. Therefore, the correct answer is (D) I and III only
Rational functions are continuous everywhere except where the denominator is 0.