Math Problem Statement
Consider the function.
f(x) =
5,
x < 1
5x,
x ≥ 1
(a)
Sketch the graph of the function.
(b)
Use the definition of a derivative to show that the function is not differentiable at
x = 1.
As
x → 1−
f '(1)
=
As
x → 1+
f '(1)
=
The limits ? are are not equal.
Solution
Let's break down the problem step by step.
Part (a): Sketch the Graph of the Function
The function is defined in a piecewise manner:
\begin{cases} 5 & \text{if } x < 1 \\ 5x & \text{if } x \geq 1 \end{cases}$$ 1. For $$x < 1$$, the function is constant at $$f(x) = 5$$. - This means the graph will be a horizontal line at $$y = 5$$ for all $$x$$ values less than 1. 2. For $$x \geq 1$$, the function is linear with a slope of 5 (i.e., $$f(x) = 5x$$). - This means the graph will be a straight line starting from the point $$(1, 5)$$ and having a slope of 5. **Graph Overview:** - The left side (for $$x < 1$$) will be a horizontal line at $$y = 5$$. - At $$x = 1$$, the function jumps to the value $$f(1) = 5$$, and for $$x > 1$$, the function increases linearly with slope 5. - There is no smooth transition at $$x = 1$$ because the function changes from a constant to a linear behavior, which might cause a corner or discontinuity. ### Part (b): Use the Definition of Derivative to Show That the Function Is Not Differentiable at $$x = 1$$ To check whether the function is differentiable at $$x = 1$$, we need to compute the left-hand and right-hand derivatives at $$x = 1$$. The derivative of a function $$f(x)$$ at a point $$x = c$$ is defined as: $$f'(c) = \lim_{h \to 0} \frac{f(c + h) - f(c)}{h}$$ #### Left-Hand Derivative at $$x = 1$$ (as $$x \to 1^-$$) For $$x < 1$$, the function is constant at $$f(x) = 5$$. So, for any value of $$h$$ approaching 0 from the left, the function value does not change. $$f'(1^-) = \lim_{h \to 0^-} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0^-} \frac{5 - 5}{h} = 0$$ So, the left-hand derivative is: $$f'(1^-) = 0$$ #### Right-Hand Derivative at $$x = 1$$ (as $$x \to 1^+$$) For $$x \geq 1$$, the function is $$f(x) = 5x$$. The derivative of $$5x$$ is simply 5. Thus, for values of $$h$$ approaching 0 from the right, we have: $$f'(1^+) = \lim_{h \to 0^+} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0^+} \frac{5(1 + h) - 5}{h} = \lim_{h \to 0^+} \frac{5h}{h} = 5$$ So, the right-hand derivative is: $$f'(1^+) = 5$$ #### Conclusion: The Function is Not Differentiable at $$x = 1$$ Since the left-hand and right-hand derivatives are not equal ($$f'(1^-) = 0$$ and $$f'(1^+) = 5$$), the function is **not differentiable** at $$x = 1$$. ### Answer for the Limits: - As $$x \to 1^-$$, $$f'(1^-) = 0$$ - As $$x \to 1^+$$, $$f'(1^+) = 5$$ The limits are not equal, which confirms that the function is not differentiable at $$x = 1$$. --- ### Do you want more details or clarifications on any part of this explanation? Here are 5 questions related to this topic: 1. How would the graph change if the function were continuous at $$x = 1$$? 2. What does the concept of differentiability mean in terms of the shape of the graph? 3. How can we prove that a function is continuous at a point using limits? 4. What happens to the derivative if the function has a sharp corner or cusp? 5. Can a function be continuous but not differentiable at a point? **Tip**: When checking differentiability at a point, always make sure to check both the left-hand and right-hand limits for the derivative. If they are not equal, the function is not differentiable at that point.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Graphing Functions
Differentiability
Limits
Derivatives
Formulas
Definition of Derivative: f'(c) = lim (h → 0) [(f(c + h) - f(c)) / h]
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
Definition of Differentiability
Limit Definition of Derivative
Suitable Grade Level
Grades 11-12 (Pre-Calculus/Calculus I)
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