Math Problem Statement
Let m be a positive constant. Consider the piecewise function:
f(x) = { m^2x - 12, x < -1
7mx, -1 <= x <= m
m / (x - 2), x > m }
(a) Use the definition of continuity to find all values of m, if any, that guarantee that f is continuous at x = -1.
(b) Use the definition of the derivative to find all values of m, if any, that guarantee that f is differentiable at x = -1.
Hint: Note that f(-1) = -m^2 - 12. Can you use your answer to part (a)?
(c) For one of the values of m that you found in part (a), determine the type of discontinuity that f has at x = m.
Solution
Let's solve the problem based on the image you've provided. We are given a piecewise function and asked to check continuity, differentiability, and discontinuity.
The piecewise function is:
\begin{cases} m^2x - 12 & \text{for } x < -1 \\ 7mx & \text{for } -1 \leq x \leq m \\ \frac{m}{x - 2} & \text{for } x > m \end{cases}$$ ### (a) **Continuity at $$x = -1$$** To determine whether $$f(x)$$ is continuous at $$x = -1$$, we need to apply the definition of continuity: A function is continuous at $$x = c$$ if: $$\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$$ For $$f(x)$$ to be continuous at $$x = -1$$, the left-hand limit ($$x \to -1^-$$) and the right-hand limit ($$x \to -1^+$$) must be equal to $$f(-1)$$. 1. **Left-hand limit ($$x \to -1^-$$):** $$\lim_{x \to -1^-} f(x) = m^2(-1) - 12 = -m^2 - 12$$ 2. **Right-hand limit ($$x \to -1^+$$):** $$\lim_{x \to -1^+} f(x) = 7m(-1) = -7m$$ 3. **Function value at $$x = -1$$:** $$f(-1) = 7m(-1) = -7m$$ For continuity, the left-hand limit must equal the right-hand limit and the function value: $$-m^2 - 12 = -7m$$ Solving this equation: $$m^2 + 7m - 12 = 0$$ Using the quadratic formula: $$m = \frac{-7 \pm \sqrt{7^2 - 4(1)(-12)}}{2(1)} = \frac{-7 \pm \sqrt{49 + 48}}{2} = \frac{-7 \pm \sqrt{97}}{2}$$ Thus, the two possible values for $$m$$ are: $$m = \frac{-7 + \sqrt{97}}{2} \quad \text{(positive root, valid as $$m$$ is positive)}$$ ### (b) **Differentiability at $$x = -1$$** A function is differentiable at a point if its derivative exists and is the same from both sides at that point. So we need to check if the derivatives from the left and right match at $$x = -1$$. The left-hand derivative is the derivative of $$m^2x - 12$$ evaluated at $$x = -1$$: $$f'_L(x) = \frac{d}{dx}(m^2x - 12) = m^2$$ The right-hand derivative is the derivative of $$7mx$$ evaluated at $$x = -1$$: $$f'_R(x) = \frac{d}{dx}(7mx) = 7m$$ For differentiability at $$x = -1$$, we require: $$m^2 = 7m$$ Solving this: $$m(m - 7) = 0$$ Thus, $$m = 7$$ (since $$m = 0$$ is not allowed as $$m$$ is positive). ### (c) **Discontinuity at $$x = m$$ for $$m = 7$$** Now, we consider the point $$x = m = 7$$. We need to check the type of discontinuity at this point by evaluating the limits from the left and right and comparing them with $$f(7)$$. 1. **Left-hand limit ($$x \to 7^-$$):** $$\lim_{x \to 7^-} f(x) = 7m = 7(7) = 49$$ 2. **Right-hand limit ($$x \to 7^+$$):** $$\lim_{x \to 7^+} f(x) = \frac{7}{7 - 2} = \frac{7}{5}$$ 3. **Function value at $$x = 7$$:** $$f(7) = 7(7) = 49$$ Since the right-hand limit $$\frac{7}{5}$$ does not equal the left-hand limit or the function value, there is a discontinuity at $$x = 7$$. This is a **jump discontinuity**. --- Would you like more details or have any further questions? Here are some additional questions related to this problem: 1. How does the quadratic formula work in solving the continuity condition? 2. What other types of discontinuities can occur in piecewise functions? 3. How do we check differentiability for more complex piecewise functions? 4. Why is the left-hand and right-hand derivative crucial for differentiability? 5. Could $$f(x)$$ be continuous at other points of the domain? **Tip:** When checking continuity, always ensure to evaluate both the limits and the function value at the point of interest to avoid missing possible discontinuities.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Continuity
Differentiability
Limits
Quadratic Equations
Formulas
Continuity: lim(x -> c^-) f(x) = lim(x -> c^+) f(x) = f(c)
Quadratic Equation: ax^2 + bx + c = 0
Definition of Derivative: f'(x) = lim(h -> 0) [f(x + h) - f(x)] / h
Theorems
Quadratic Formula
Continuity and Differentiability Theorem
Suitable Grade Level
Grades 11-12
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