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Continuity And Differentiability Homework Answers

Ok, you can see easily that $f(x)$ is continuous at $x=1$ because $f(1)=-2$ and $$\lim_{x\to 1^-}f(x)=\lim_{x\to 1^+}f(x)=-2$$

A function $f(x)$ is continuous at $x=c$ if it is defined at $x=c$ and $\lim_{x\to c}=f(c)$

Now we differentiate the function in each piece. If the result is continuous then the function is differentiable

$f'(x)=\begin{cases} 2x, &x<1\\ 0, &x=1\\ \dfrac{1}{2\sqrt{x-1}}, &x>1\end{cases}$

we have

$$\lim_{x\to 1^-}f'(x)=2;\;\lim_{x\to 1^+}f'(x)=+\infty$$ the limit doesn't even exist! So the function is not differentiable at $x=1$.


The derivative which tends to $+\infty$ has an interesting geometric meaning. It means that the tangent to the graph of the function is vertical at that point. As you can see in the picture below, the graph has two tangents at point $(1,\;-2)$, which means it is not differentiable, indeed.

Hope this is useful


answered Dec 3 '17 at 20:18

Continuity & Differentiability
miscellaneous on-line topics for
Calculus Applied to the Real World
Part A: Continuity



Part A: Continuity

Note To understand this topic, you will need to be familiar with limits, as discussed in the chapter on derivatives in Calculus Applied to the Real World. If you like, you can review the topic summary material on limits or, for a more detailed study, the on-line tutorial on limits.

To begin, we recall the definition (see Section 6 in the derivatives chapter in Calculus Applied to the Real World.) of what it means for a function to be continuous at a point or on a subset of its domain.

Continuous at a Point
The function $f$ is continuous at the point a in its domain if:
    $f(x)$   exists;
    $f(x) = f(a)$
If $f$ is not continuous at $a,$ we say that $f$ is discontinuous at $a.$ Note
If the point $a$ is not in the domain of $f,$ we do not talk about whether or not $f$ is continuous at $a.$ Continuous on a Subset of the Domain
The function $f$ is continuous on the subset $S$ of its domain if it continuous at each point of $S.$ Examples
1. All closed form functions are continuous on their (whole) domain. A closed-form function is any function that can be obtained by combining constants, powers of $x,$ exponential functions, radicals, logarithms, and trigonometric functions (and some other functions we shall not encounter here) into a single mathematical formula by means of the usual arithmetic operations and composition of functions. Examples of closed form functions are:
    $3x^2 - x + 1,$     $\frac{2x}{x + 3},$     and     $e^{x^{2-1}}.$

2. The function $f(x) = 1/x,$ also of closed form, is continuous at every point of its domain. (Note that 0 is not a point of the domain of $f,$ so we don't discuss what it might mean to be continous or discontinuous there.)

3. The function

    $f(x)   =$ $-1$
      if $x  ≤  2$
    if $x  >  2$
is not a closed-form function (since we need two algebraic formulas to specify it). Moreover, it is not continuous at $x = 2,$ since $lim_{x→2}f(x)$ does not exist.

Example 1 Recognizing Points of Discontinuity in a Graph

Let f have the graph shown below.

At which points of the domain of $f$ is the function $f$ discontinuous?


According to the definition, $f$ can fail to be continuous at a point $a$ of its domain if either:
    $f(x)$   does not exist, or
    $f(x)$   exists, but $≠ f(a)$
Looking at the figure, we see that the possible points where things might go wrong are at $x = -1, 0, 1,$ and $2.$ Let us look at these points one-at-a-time. $x = -1:$ Notice that $x = -1$ is a point of the domain of $f,$ and $lim_{x→-1}f(x)$ does not exist, because the left- and right-habd limits disagree. Therefore, $f$ has a discontinuity at $x = -1.$ $x = 0:$ From the graph, we see that $f(0)$ is not defined. Therefore, $0$ is not in the domain of $f,$ so we cannot say that $f$ has a discontinuity at $x = 0.$ (Some authors would say that $f$ is discontinuous at $x = 0,$ but we don't consider such points...) $x = 1:$ At the point where $x = 1,$ the limit does exist, however,
    $f(x) = 1,$   but
    $f(1) = 2$
Thus, since the limit at $1$ does not equal the value of the function, we have a discontinuity at $x = 1.$ $x = 2:$ Although there is a cusp in the graph at $x = 2,$ you will see that the limit at $x = 2$ exists, and agrees with $f(2).$ Therefore, $f$ is continuous at $x = 2.$ Thus, the only points of discontinuity in the domain of $f$ occur at $x = -1$ and $x =1.$

Here is a more interactive example.

Example 2 Recognizing Points of Discontinuity in a Graph

Let $f$ have the graph shown below.

The next examples discuss functions specified algebraically rather than geometrically.

Example 3 Identifying Points of Discontinuity in a Non-Closed Form Function

Let $f$ be specified as
    $f(x)   =$ $x^2+4$
    $x + 1$
    $x^2 + 1$
      if $x  ≤  0$
    if $0  <  x  ≤  1$
    if $x  >  1$
Then, the only conceivable points of discontinuity occur at $x = 0$ and $x = 1.$ (At all other points in the domain, $f$ can be specified by a single closed-form formula; for example, near $x = 0.5,$ $f$ can be specified as $x + 1$). Looking near $x = 0,$ we get
    $f(x)   =   4$         Substitute $x = 0$ in the first (closed-form) formula
    $f(x)   =   1$         Substitute $x = 0$ in the second (closed-form) formula
Since these limits disagree, $lim_{x→0}f(x)$ does not exist, and so the function is discontinuous at $x = 0.$ If you now compute the left- and right limits at $x = 1,$ on the other hand, you will find they agree and equal $2,$ which is also the value of $f(1).$ Thus, $f$ is continuous at $x = 1,$ and so the only point of discontinuity is $x = 0.$

Example 4 Fixing up a Function

Let $f$ be given by
    $f(x)   =$ $x^2 - x + 1$
    $kx^2 + 1$
      if $x  ≤  1$
    if $x  >  1$
For which value(s) of $k$ is $f$ continuous on its domain?


Since the only place where something can go wrong is at $x = 1,$ we look at the left- and right limits, as well as the value of the function.
    $f(x)   =   1$
    $f(x)   =   k+1$
    $f(1) = 1$
To make the left- and right limits agree, we must make $k = 1$ (and this also makes $f(1)$ agree with those limits). Thus, we take $k = 1.$

You can now either go on and try those exercises that deal with continuity in the exercise set for this topic, or first finish the on-line text by going to Part B: Differentiability.

Copyright © 1999 StefanWaner and Steven R. Costenoble