Elishipa Mugo
10 hours ago
Find a and b: -2x if x < 1, b - ax^2 if 1 < x < 4, -16x if x > 4.
In contuinity of a function
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ChatGPT
10 hours ago
In order to find the values of a and b, we need to consider the different intervals given for x.
For x < 1:
-2x = -2(1) = -2
For 1 < x < 4:
b - ax^2 = b - a(2^2) = b - 4a
For x > 4:
-16x = -16(4) = -64
Since the function is continuous, the values of the function at x = 1 and x = 4 must be equal. Therefore, we can set the values of the function at x = 1 and x = 4 equal to each other:
-2 = b - 4a
-64 = b - 4a
Solving these two equations simultaneously, we get:
b - 4a = -2
b - 4a = -64
Subtracting the first equation from the second, we get:
-62 = -62
This means that the system of equations is consistent and has infinitely many solutions. Therefore, we cannot determine unique values for a and b based on the information given.
For x < 1:
-2x = -2(1) = -2
For 1 < x < 4:
b - ax^2 = b - a(2^2) = b - 4a
For x > 4:
-16x = -16(4) = -64
Since the function is continuous, the values of the function at x = 1 and x = 4 must be equal. Therefore, we can set the values of the function at x = 1 and x = 4 equal to each other:
-2 = b - 4a
-64 = b - 4a
Solving these two equations simultaneously, we get:
b - 4a = -2
b - 4a = -64
Subtracting the first equation from the second, we get:
-62 = -62
This means that the system of equations is consistent and has infinitely many solutions. Therefore, we cannot determine unique values for a and b based on the information given.
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Google Gemini AI
10 hours ago
Given:
$$f(x) = \begin{cases} -2x & \text{if } x < 1 \\\ b - ax^2 & \text{if } 1 < x < 4 \\\ -16x & \text{if } x > 4 \end{cases}$$
Condition 1: Continuity at x = 1
For the function to be continuous at x = 1, the left-hand limit and right-hand limit must be equal to the value of the function at x = 1.
Left-hand limit (x → 1⁻):
$$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (-2x) = -2(1) = -2$$
Right-hand limit (x → 1⁺):
$$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (b - ax^2) = b - a(1)^2 = b - a$$
Value of the function at x = 1:
$$f(1) = b - a(1)^2 = b - a$$
For continuity at x = 1:
$$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$$
$$-2 = b - a$$
Condition 2: Differentiability at x = 1
For the function to be differentiable at x = 1, the left-hand derivative and right-hand derivative must exist and be equal.
Left-hand derivative (x → 1⁻):
$$f'(x) = \lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^-} \frac{(-2(1+h)) - (-2)}{h} = -2$$
Right-hand derivative (x → 1⁺):
$$f'(x) = \lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^+} \frac{(b - a(1+h)^2) - (b - a)}{h}$$
$$= \lim_{h \to 0^+} \frac{-2ah - ah^2}{h} = -2a$$
For differentiability at x = 1:
$$f'(1^-) = f'(1^+)$$
$$-2 = -2a$$
Solving this equation, we get:
$$a = 1$$
Substituting this value of a back into the equation from the first condition:
$$-2 = b - 1$$
$$b = -1$$
Therefore, a = 1 and b = -1.
$$f(x) = \begin{cases} -2x & \text{if } x < 1 \\\ b - ax^2 & \text{if } 1 < x < 4 \\\ -16x & \text{if } x > 4 \end{cases}$$
Condition 1: Continuity at x = 1
For the function to be continuous at x = 1, the left-hand limit and right-hand limit must be equal to the value of the function at x = 1.
Left-hand limit (x → 1⁻):
$$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (-2x) = -2(1) = -2$$
Right-hand limit (x → 1⁺):
$$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (b - ax^2) = b - a(1)^2 = b - a$$
Value of the function at x = 1:
$$f(1) = b - a(1)^2 = b - a$$
For continuity at x = 1:
$$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$$
$$-2 = b - a$$
Condition 2: Differentiability at x = 1
For the function to be differentiable at x = 1, the left-hand derivative and right-hand derivative must exist and be equal.
Left-hand derivative (x → 1⁻):
$$f'(x) = \lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^-} \frac{(-2(1+h)) - (-2)}{h} = -2$$
Right-hand derivative (x → 1⁺):
$$f'(x) = \lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^+} \frac{(b - a(1+h)^2) - (b - a)}{h}$$
$$= \lim_{h \to 0^+} \frac{-2ah - ah^2}{h} = -2a$$
For differentiability at x = 1:
$$f'(1^-) = f'(1^+)$$
$$-2 = -2a$$
Solving this equation, we get:
$$a = 1$$
Substituting this value of a back into the equation from the first condition:
$$-2 = b - 1$$
$$b = -1$$
Therefore, a = 1 and b = -1.
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