Math Problem Statement
Solution
To solve the problem presented in the image, let's analyze the given information:
The function is defined piecewise as:
\begin{cases} 3e^{2x}, & x < k \\ 2e^{3x}, & x \geq k \end{cases}$$ ### Part (a) Determine $$k$$ such that $$f$$ is continuous. For $$f(x)$$ to be continuous, the two pieces of $$f(x)$$ must meet at $$x = k$$. This requires: $$\lim_{x \to k^-} f(x) = \lim_{x \to k^+} f(x) = f(k).$$ 1. **Left-hand limit (as $$x \to k^-$$):** $$\lim_{x \to k^-} f(x) = 3e^{2k}.$$ 2. **Right-hand limit (as $$x \to k^+$$):** $$\lim_{x \to k^+} f(x) = 2e^{3k}.$$ For continuity, these limits must be equal: $$3e^{2k} = 2e^{3k}.$$ Divide through by $$e^{2k}$$ (valid since $$e^{2k} > 0$$): $$3 = 2e^{k}.$$ Solve for $$k$$: $$e^{k} = \frac{3}{2}.$$ Taking the natural logarithm: $$k = \ln\left(\frac{3}{2}\right).$$ Thus, $$k = \ln\left(\frac{3}{2}\right)$$ ensures $$f(x)$$ is continuous. --- ### Part (b) Show that no $$k$$ can make $$f$$ differentiable. For $$f(x)$$ to be differentiable at $$x = k$$, the derivatives of the two pieces must also match at $$x = k$$. Compute the derivatives of each piece: 1. **Derivative for $$x < k$$:** $$f'(x) = \frac{d}{dx}[3e^{2x}] = 6e^{2x}.$$ At $$x = k$$, this becomes: $$f'(k^-) = 6e^{2k}.$$ 2. **Derivative for $$x \geq k$$:** $$f'(x) = \frac{d}{dx}[2e^{3x}] = 6e^{3x}.$$ At $$x = k$$, this becomes: $$f'(k^+) = 6e^{3k}.$$ For $$f(x)$$ to be differentiable at $$x = k$$, we must have: $$f'(k^-) = f'(k^+).$$ This requires: $$6e^{2k} = 6e^{3k}.$$ Simplify by dividing through by $$6e^{2k}$$: $$1 = e^{k}.$$ Taking the natural logarithm: $$k = 0.$$ However, if $$k = 0$$, substituting into the continuity condition: $$3e^{2k} = 2e^{3k} \implies 3 = 2,$$ which is a contradiction. Thus, no value of $$k$$ can make $$f(x)$$ differentiable. --- ### Final Answers: (a) $$k = \ln\left(\frac{3}{2}\right)$$ ensures $$f(x)$$ is continuous. (b) No $$k$$ can make $$f(x)$$ differentiable. --- Would you like a detailed explanation of any step, or additional examples? ### Related Questions: 1. What does continuity imply about the relationship between limits and function values? 2. How does differentiability depend on continuity? 3. Can a piecewise function ever be differentiable at a boundary point? 4. What happens when the slopes of two pieces of a function do not match at a point? 5. How would the analysis change if the function were defined differently? ### Tip: Always verify continuity before checking differentiability; differentiability implies continuity, but not vice versa!Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Continuity of functions
Differentiability of piecewise functions
Exponential functions
Formulas
Continuity condition: lim(x→k⁻) f(x) = lim(x→k⁺) f(x) = f(k)
Differentiability condition: f'(k⁻) = f'(k⁺)
Exponential rules: e^(a+b) = e^a * e^b
Theorems
A function is differentiable at a point only if it is continuous at that point.
Limits and continuity principles.
Suitable Grade Level
High School (Grades 11-12) or University Introductory Level
Related Recommendation
Continuity and Differentiability of a Piecewise Function at a Point
Continuity and Differentiability of a Piecewise Function at a Point
Determine Values of k and p for Continuity and Differentiability of a Piecewise Function at x = 1
Find k for Continuity of f(x) at x=2 for a Piecewise Function
Finding k for Continuity in a Piecewise Function