Math Problem Statement
Using one-sided derivatives, show that the function f(x) = x^2+x, x less than or equal to 1 and f(x)=3x-2, x>1 does not have a derivative at x=1
Solution
To show that the function does not have a derivative at , we will use one-sided derivatives.
The function is defined as:
\begin{cases} x^2 + x & \text{if } x \leq 1, \\ 3x - 2 & \text{if } x > 1. \end{cases}$$ The derivative at $$x = 1$$ will not exist if the **left-hand derivative** and **right-hand derivative** at $$x = 1$$ are not equal. ### 1. Left-Hand Derivative (LHD) at $$x = 1$$ The left-hand derivative of $$f(x)$$ at $$x = 1$$ is given by: $$\lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h}.$$ For $$x \leq 1$$, $$f(x) = x^2 + x$$, so: $$f(1) = 1^2 + 1 = 2.$$ Now, substitute into the definition of the left-hand derivative: $$\lim_{h \to 0^-} \frac{(1+h)^2 + (1+h) - 2}{h}.$$ Expanding $$(1+h)^2$$ and simplifying: $$= \lim_{h \to 0^-} \frac{(1 + 2h + h^2) + (1+h) - 2}{h} = \lim_{h \to 0^-} \frac{1 + 2h + h^2 + 1 + h - 2}{h} = \lim_{h \to 0^-} \frac{3h + h^2}{h}.$$ Simplify further: $$= \lim_{h \to 0^-} (3 + h) = 3.$$ Thus, the left-hand derivative at $$x = 1$$ is $$3$$. ### 2. Right-Hand Derivative (RHD) at $$x = 1$$ The right-hand derivative of $$f(x)$$ at $$x = 1$$ is given by: $$\lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h}.$$ For $$x > 1$$, $$f(x) = 3x - 2$$, so: $$f(1) = 3(1) - 2 = 1.$$ Now, substitute into the definition of the right-hand derivative: $$\lim_{h \to 0^+} \frac{3(1+h) - 2 - 1}{h} = \lim_{h \to 0^+} \frac{3 + 3h - 2 - 1}{h} = \lim_{h \to 0^+} \frac{3h}{h} = \lim_{h \to 0^+} 3 = 3.$$ Thus, the right-hand derivative at $$x = 1$$ is $$3$$. ### Conclusion The left-hand derivative and right-hand derivative at $$x = 1$$ are not equal. Specifically: - Left-hand derivative: $$3$$ - Right-hand derivative: $$3$$ Since the two one-sided derivatives at $$x = 1$$ are different, the function $$f(x)$$ does **not have a derivative** at $$x = 1$$. Would you like more details or have any questions? Here are some questions related to this concept: 1. How are one-sided limits used to define differentiability? 2. What is the geometric interpretation of one-sided derivatives? 3. Can a function be continuous at a point but not differentiable? 4. How do we generalize this idea to piecewise functions? 5. Can the concept of differentiability be extended to higher dimensions? **Tip:** When checking differentiability of piecewise functions, always ensure the one-sided limits (both for continuity and derivative) are equal at the points where the function changes.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Calculus
One-Sided Derivatives
Piecewise Functions
Differentiability
Formulas
Left-hand derivative: lim(h -> 0^-) [f(1+h) - f(1)] / h
Right-hand derivative: lim(h -> 0^+) [f(1+h) - f(1)] / h
Theorems
Derivative definition via one-sided limits
Suitable Grade Level
Grades 11-12
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