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Magazine Chapter 2: Problem 47
Show that $$f(x)=\left\\{\begin{array}{ll}{x^{2}+1,} & {x \leq 1} \\ {2 x,} & {x>1}\end{array}\right.$$ is continuous and differentiable at \(x=1 .\) Sketch the graph of \(f .\)
Short Answer
Expert verified
The function \(f(x)\) is continuous and differentiable at \(x=1\). The graph is a parabola for \(x \leq 1\) and a line for \(x > 1\).
Step by step solution
01
Check Continuity at x = 1
To show that the function is continuous at \(x=1\), we must prove that \(\lim_{{x \to 1^-}} f(x) = \lim_{{x \to 1^+}} f(x) = f(1)\). First evaluate \(f(1)\). Since \(x=1\), use the first part of the piecewise function: \(f(1) = 1^2 + 1 = 2\). Next, compute the left-hand limit: \[\lim_{{x \to 1^-}} f(x) = \lim_{{x \to 1^-}} (x^2+1) = 1^2 + 1 = 2\]. Now compute the right-hand limit: \(\lim_{{x \to 1^+}} f(x) = \lim_{{x \to 1^+}} 2x = 2 \times 1 = 2\). Since both limits and \(f(1)\) are equal, \(f(x)\) is continuous at \(x=1\).
02
Check Differentiability at x = 1
For differentiability at \(x=1\), the left-hand derivative must equal the right-hand derivative. Compute the left-hand derivative: \(\lim_{{h \to 0^-}} \frac{(1+h)^2 + 1 - 2}{h} = \lim_{{h \to 0^-}} \frac{h^2 + 2h}{h} = \lim_{{h \to 0^-}} (h + 2) = 2\). Now the right-hand derivative: \(\lim_{{h \to 0^+}} \frac{2(1+h) - 2}{h} = \lim_{{h \to 0^+}} \frac{2 + 2h - 2}{h} = \lim_{{h \to 0^+}} 2 = 2\). Since both derivatives are equal, \(f(x)\) is differentiable at \(x=1\).
03
Sketch the Graph of f(x)
To graph \(f(x)\), plot the part \(x^2+1\) for \(x \leq 1\) and \(2x\) for \(x>1\). At \(x=1\), there is no jump since the function is continuous, and smoothly transitions from one part to the other. The curve \(x^2+1\) starts from (0, 1) and moves up to (1, 2). The line \(2x\) starts from (1, 2) and extends upwards. The graph is a parabola up to \(x=1\) and a line past \(x=1\).
04
Conclusion
We have shown that \(f(x)\) is continuous and differentiable at \(x=1\) by evaluating the limits and derivatives at this point, and graphed the function across its domain.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Piecewise Function
A piecewise function is a mathematical expression defined by different expressions depending on the value of the independent variable. In this exercise, the function \( f(x) \) is defined differently for values of \( x \) less than or equal to 1, and for \( x \) greater than 1.
Piecewise functions are common in scenarios where a single formula is not sufficient to describe the behavior of a function across its entire domain.
Piecewise functions are common in scenarios where a single formula is not sufficient to describe the behavior of a function across its entire domain.
- For \( x \leq 1 \), \( f(x) \) is given by \( x^2 + 1 \).
- For \( x > 1 \), \( f(x) \) changes to a linear expression \( 2x \).
Differentiability
Differentiability is a property that indicates whether a function has a derivative, which is a measure of how the function changes at any point. To prove differentiability at a specific point for a piecewise function, we must ensure that both the left-hand and right-hand derivatives exist and are equal at that point.
In this problem, we are looking at \( x = 1 \):
This smooth transition signifies a lack of sharp corners or cusps at \( x = 1 \), which confirms the function's neat staircase-like ascent from one rule to another.
In this problem, we are looking at \( x = 1 \):
- For the left-hand derivative, we considered the \( h \to 0^- \) limit from the \( x^2+1 \) expression, which led to \( 2 \).
- For the right-hand derivative, we examined the \( h \to 0^+ \) limit from \( 2x \), resulting in \( 2 \) as well.
This smooth transition signifies a lack of sharp corners or cusps at \( x = 1 \), which confirms the function's neat staircase-like ascent from one rule to another.
Limits
The concept of limits is central to calculus and helps define both the continuity and differentiability of piecewise functions. A limit describes the value that a function approaches as the input (in this case, \( x \)) tends towards a certain point.
The exercise required checking the limits approaching \( x = 1 \) from both sides:
The limit analysis reveals that there are no jumps or breaks at the transition point - a vital check when working with piecewise functions.
The exercise required checking the limits approaching \( x = 1 \) from both sides:
- The left-hand limit was determined by \( x^2 + 1 \) as \( x \rightarrow 1^- \) and equated to \( 2 \).
- The right-hand limit was determined using \( 2x \) as \( x \rightarrow 1^+ \), also equating to \( 2 \).
The limit analysis reveals that there are no jumps or breaks at the transition point - a vital check when working with piecewise functions.
Derivatives
A derivative expresses the rate of change of a function with respect to its variable, often thought of as a curve's slope at any given point. To assess differentiability of a piecewise function at a point of transition, it's crucial that both the left and right derivatives converge to a common value at that point.
Derivatives like these allow us to make precise statements about the behavior and changes of functions across their domains, particularly at critical points like where rules of a piecewise function switch.
- The left-hand derivative at \( x = 1 \) was found by evaluating the limit \( h \to 0^- \) for the expression \( x^2 + 1 \), resulting in \( 2 \).
- Similarly, the right-hand derivative was calculated as \( h \to 0^+ \) using \( 2x \), also giving \( 2 \).
Derivatives like these allow us to make precise statements about the behavior and changes of functions across their domains, particularly at critical points like where rules of a piecewise function switch.
