Wize High School Grade 12 Calculus Textbook > Rate of Change

Continuity
f(x)f(x)f(x) is continuous at x=a if lim⁡x→af(x)=f(a)\displaystyle\lim_{x\to a}f\left(x\right)=f\left(a\right)x→alim​f(x)=f(a).
f(x)f\left(x\right)f(x)is discontinuous at x=a if:
lim⁡x→af(x)\displaystyle\lim_{x\to a}f\left(x\right)x→alim​f(x)does not exist or
lim⁡x→af(x)≠f(a)\displaystyle\lim_{x\to a}f\left(x\right)\ne f\left(a\right)x→alim​f(x)=f(a)or
f(x)f\left(x\right)f(x)is undefined at 𝑥 = 𝑎
A function is continuous on its domain if it is continuous at every point on its domain.


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Types of Discontinuities

Note: 
Polynomials are continuous for all real numbers
Rational functions f(x)g(x)\frac{f\left(x\right)}{g\left(x\right)}g(x)f(x)​ is continuous where f(x)f\left(x\right)f(x) and g(x)g\left(x\right)g(x) are continuous except for when g(x)=0g\left(x\right)=0g(x)=0
Composition of continuous functions are also continuous
Limits can "flow" through continuous functions

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Example: Discontinuity
Find all value(s) of xxx at which the following functions are discontinuous. State the type of discontinuity.

a) r(x)=3(x−1)2+x−4r(x)=3(x-1)^2+x-4r(x)=3(x−1)2+x−4

This function is continuous everywhere.

b) f(x)=1xf\left(x\right)=\frac{1}{x}f(x)=x1​

This is a rational function, it is discontinuous at x=0\boxed{x=0}x=0​.
The function is not defined at x=0x=0x=0
lim⁡x→0−f(x)=−∞\displaystyle \lim_{x\to0^-}f(x)=-\inftyx→0−lim​f(x)=−∞ and lim⁡x→0+f(x)=∞\displaystyle \lim_{x\to0^+}f(x)=\inftyx→0+lim​f(x)=∞, so lim⁡x→0f(x)\displaystyle \lim_{x\to0}f(x)x→0lim​f(x) DNE
This is an infinite discontinuity.

c) g(x)=x2−25x+5\displaystyle g\left(x\right)=\frac{x^2-25}{x+5}g(x)=x+5x2−25​

This is a rational function, it is discontinuous at x=−5\boxed{x=-5}x=−5​.
The function is not defined at x=−5x=-5x=−5
Factoring the numerator and denominator: g(x)=(x+5)(x−5)x+5=(x+5)(x+5)(x−5)g\left(x\right)=\frac{\left(x+5\right)\left(x-5\right)}{x+5}=\frac{\left(x+5\right)}{\left(x+5\right)}\left(x-5\right)g(x)=x+5(x+5)(x−5)​=(x+5)(x+5)​(x−5), so the function is continuous and equals y=x−5y=x-5y=x−5 everywhere else except for at x=−5x=-5x=−5
This is a point discontinuity.

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d) h(x)=2x+1x2+3x+2\displaystyle h\left(x\right)=\frac{2x+1}{x^2+3x+2}h(x)=x2+3x+22x+1​

This is a rational function h(x)=2x+1(x+2)(x+1)\displaystyle h(x)=\frac{2x+1}{(x+2)(x+1)}h(x)=(x+2)(x+1)2x+1​, it is discontinuous at x=−2  and  x=−1\boxed{x=-2\ \text{ and }\ x=-1}x=−2  and  x=−1​
The function is not defined at x=−2x=-2x=−2 or x=−1x=-1x=−1
lim⁡x→−2−h(x)=−∞\displaystyle \lim_{x\to-2^-}h(x)=-\inftyx→−2−lim​h(x)=−∞ and lim⁡x→−2+h(x)=∞\displaystyle \lim_{x\to-2^+}h(x)=\inftyx→−2+lim​h(x)=∞, so lim⁡x→−2h(x)\displaystyle \lim_{x\to-2}h(x)
x→−2lim​h(x) DNE
This is an infinite discontinuity.

e) p(x)={x,if x<03x+2,if x≥0p\left(x\right)=
\begin{cases}
x,&\text{if }x<0\\
3x+2,&\text{if }x\ge0
\end{cases}p(x)={x,3x+2,​if x<0if x≥0​

This is a piecewise function, each of the "pieces" are continuous. We need to check the "connection" point.
p(0)=3(0)+2=2p\left(0\right)=3\left(0\right)+2=2p(0)=3(0)+2=2, so the function is defined at x=0x=0x=0
lim⁡x→0−p(x)=lim⁡x→0−x=0\displaystyle \lim_{x\to0^-}p(x)= \lim_{x\to0^-}x=0x→0−lim​p(x)=x→0−lim​x=0 and lim⁡x→0+p(x)=lim⁡x→0+3x+2=2\displaystyle \lim_{x\to0^+}p(x)= \lim_{x\to0^+}3x+2=2x→0+lim​p(x)=x→0+lim​3x+2=2, so lim⁡x→0p(x)\displaystyle \lim_{x\to0}p(x)x→0lim​p(x) DNE
So, the function is discontinuous at x=0\boxed{x=0}x=0​
This is a jump discontinuity.

f) q(x)=x−5q(x)=\sqrt {x-5}q(x)=x−5​

The function is not defined when x−5<0x-5<0x−5<0, so it is discontinuous when x<5\boxed {x<5}x<5​

Practice: Discontinuity From Graphs
The following is the graph of y=f(x)y=f\left(x\right)y=f(x). On what interval is g(x)=1f(x)−1g\left(x\right)=\frac{1}{f\left(x\right)-1}g(x)=f(x)−11​ discontinuous?

x=4x=4x=4 only

x=0, 4x=0,\ 4x=0, 4 only

x=0,  4,  7.25,  7.75,  10.25x=0,\ \ 4,\ \ 7.25,\ \ 7.75,\ \ 10.25x=0,  4,  7.25,  7.75,  10.25 only

x=2, 4, 7, 8, 10x=2,\ 4,\ 7,\ 8,\ 10x=2, 4, 7, 8, 10 only

x=4,  7.4,  7.6,  10.4x=4,\ \ 7.4,\ \ 7.6,\ \ 10.4x=4,  7.4,  7.6,  10.4 only

I don't know

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Example: Continuity of Piecewise Functions
Where is the function 𝑓 continuous?
f(x)={0if  x≤0x2if  0<x≤15−xif  x>1f\left(x\right)=
\begin{cases}
0 & \text{if }\space x\le0\\
x^2&\text{if }\space 0<x\le1 \\
\sqrt{5-x}& \text{if }\space x>1

\end{cases}f(x)=⎩⎨⎧​0x25−x​​if  x≤0if  0<x≤1if  x>1​

1. We need to check each piece to see if they're continuous:
000 is continuous everywhere
x2x^2x2 is continuous everywhere
5−x\sqrt{5-x}5−x​ is discontinuous when x>5x>5x>5. This piece of the function has the domain x>1x>1x>1. The overlapping part is x>5x>5x>5  (Discontinuous points so far: x > 5)

2. We now check the connecting points:
x=0x=0x=0
lim⁡x→0−f(x)=lim⁡x→0−0=0\displaystyle\lim_{x\rightarrow0^-}f\left(x\right)=\lim_{x\rightarrow0^-}0=0x→0−lim​f(x)=x→0−lim​0=0
lim⁡x→0+f(x)=lim⁡x→0+x2=0\displaystyle\lim_{x\rightarrow0^+}f\left(x\right)=\lim_{x\rightarrow0^+}x^2=0x→0+lim​f(x)=x→0+lim​x2=0
f(0)=0f\left(0\right)=0f(0)=0
Since the left limit, right limit, and function at that point all equal, the function is continuous at x=0x=0x=0

x=1x=1x=1
lim⁡x→1−f(x)=lim⁡x→1−x2=1\displaystyle\lim_{x\rightarrow1^-}f\left(x\right)=\lim_{x\rightarrow1^-}x^2=1x→1−lim​f(x)=x→1−lim​x2=1

lim⁡ x→1+f(x)=lim⁡x→1+5−x=4=2\displaystyle\lim\ _{x\rightarrow1^+}f\left(x\right)=\lim_{x\rightarrow1^+}\sqrt{5-x}=\sqrt{4}=2lim x→1+​f(x)=x→1+lim​5−x​=4​=2
Since the left and right hand limits don't equal, the function is discontinuous at x=1x=1x=1 
(Discontinuous point: x = 1)

Therefore, the function is discontinuous when x=1x=1x=1 and x>5x>5x>5.
So, the function is continuous when x<1 or  1<x≤5x<1\ or\ \ 1<x\le5x<1 or  1<x≤5.
Another way to write this is (−∞,1)∪(1,5]\left(-\infty,1\right)\cup\left(1,5\right](−∞,1)∪(1,5].

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Example: Continuity of Piecewise Functions
For what value(s) of the constant kkk is the function f(x)={x−k,if x≤1kx2+k−8,if x>1f(x)=\begin{cases}
x-k,&\text{if } x\le1\\
kx^2+k-8,&\text{if }x>1
\end{cases}f(x)={x−k,kx2+k−8,​if x≤1if x>1​ continuous everywhere?

Each piece of the function is a regular polynomial, so they are continuous on their domains, meaning that
x−kx-kx−k is continuous on x≤1x\le1x≤1 and
kx2+k−8kx^2+k-8kx2+k−8 is continuous on x>1x>1x>1
We just have to make sure that the function is continuous at the connecting point x=1x=1x=1:
f(1)=1−kf\left(1\right)=1-kf(1)=1−k
lim⁡x→1−f(x)=lim⁡x→1−x−k=1−k\lim_{x\to1^-}f\left(x\right)=\lim_{x\to1^-}x-k=1-klimx→1−​f(x)=limx→1−​x−k=1−k
lim⁡x→1+f(x)=lim⁡x→1+kx2+k−8=2k−8\lim_{x\to1^+}f\left(x\right)=\lim_{x\to1^+}kx^2+k-8=2k-8limx→1+​f(x)=limx→1+​kx2+k−8=2k−8
We need all 3 of these expressions to equal one another:
1−k=2k−81-k=2k-81−k=2k−8
9=3k9=3k9=3k
k=3k=3k=3

Therefore, the function is continuous everywhere if k=3k=3k=3

Practice: Continuity of Piecewise Functions
Find the values of 𝑎 and 𝑏 that make 𝑓 continuous everywhere
f(x)={x2−2if  x<2axif  2≤x<32x−bif  x≥3f\left(x\right)=
\begin{cases}
x^2-2 & \text{if }\space x<2\\
ax & \text{if }\space 2\le x<3\\
2x-b & \text{if }\space x\ge3
\end{cases}f(x)=⎩⎨⎧​x2−2ax2x−b​if  x<2if  2≤x<3if  x≥3​

a=

b=

I don't know

Practice: Continuous Functions
Suppose that f(x)f\left(x\right)f(x) and g(x)g\left(x\right)g(x) are continuous functions on the interval −3<x<2-3<x<2−3<x<2.

If g(1)=−3g\left(1\right)=-3g(1)=−3 and lim⁡x→1(4x+g(x)−g(x)xf(x))=2\displaystyle\lim_{x\to1}\left(4x+g\left(x\right)-\frac{g\left(x\right)}{x{f\left(x\right)}}\right)=2x→1lim​(4x+g(x)−xf(x)g(x)​)=2, determine the value of f(1)f\left(1\right)f(1).

f(1)=

I don't know

Extra Practice

Piecewise Functions

Practice: Piecewise Functions
What value of 'a' would make the following piecewise function continuous?
f(x)={x2+10;x<1−x+7a;x≥1f(x)=\begin{cases}
x^2+10;&x<1\\
-x+7a;&x\geq1
\end{cases}f(x)={x2+10;−x+7a;​x<1x≥1​

Piecewise Functions

Practice: Piecewise Functions
What value of 'a' would make the following piecewise function continuous?
f(x)={−2x2+7;x<−43x+a;x≥−4f(x)=\begin{cases}
-2x^2+7;&x<-4\\
3x+a;&x\geq-4
\end{cases}f(x)={−2x2+7;3x+a;​x<−4x≥−4​

Continuity and Piecewise Functions

What values of aaa and bbb make the following function continuous? 
f(x)={ax2−4;x<−2ax+b;−2≤x≤3ax2+bx+10;x>3f(x)=
\begin{cases}
ax^2-4;&&x<-2\\\\
ax+b;&&-2\leq{}x\leq3\\\\
ax^2+bx+10;&&x>3
\end{cases}f(x)=⎩⎨⎧​ax2−4;ax+b;ax2+bx+10;​​x<−2−2≤x≤3x>3​