1.4 Inverse functions  (Page 3/11)

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Sketching graphs of inverse functions

For the graph of $f$ in the following image, sketch a graph of ${f}^{-1}$ by sketching the line $y=x$ and using symmetry. Identify the domain and range of ${f}^{-1}.$

Reflect the graph about the line $y=x.$ The domain of ${f}^{-1}$ is $\left[0,\infty \right).$ The range of ${f}^{-1}$ is $\left[-2,\infty \right).$ By using the preceding strategy for finding inverse functions, we can verify that the inverse function is ${f}^{-1}\left(x\right)={x}^{2}-2,$ as shown in the graph.

Sketch the graph of $f\left(x\right)=2x+3$ and the graph of its inverse using the symmetry property of inverse functions.

Restricting domains

As we have seen, $f\left(x\right)={x}^{2}$ does not have an inverse function because it is not one-to-one. However, we can choose a subset of the domain of $f$ such that the function is one-to-one. This subset is called a restricted domain    . By restricting the domain of $f,$ we can define a new function $g$ such that the domain of $g$ is the restricted domain of $f$ and $g\left(x\right)=f\left(x\right)$ for all $x$ in the domain of $g.$ Then we can define an inverse function for $g$ on that domain. For example, since $f\left(x\right)={x}^{2}$ is one-to-one on the interval $\left[0,\infty \right),$ we can define a new function $g$ such that the domain of $g$ is $\left[0,\infty \right)$ and $g\left(x\right)={x}^{2}$ for all $x$ in its domain. Since $g$ is a one-to-one function, it has an inverse function, given by the formula ${g}^{-1}\left(x\right)=\sqrt{x}.$ On the other hand, the function $f\left(x\right)={x}^{2}$ is also one-to-one on the domain $\left(\text{−}\infty ,0\right].$ Therefore, we could also define a new function $h$ such that the domain of $h$ is $\left(\text{−}\infty ,0\right]$ and $h\left(x\right)={x}^{2}$ for all $x$ in the domain of $h.$ Then $h$ is a one-to-one function and must also have an inverse. Its inverse is given by the formula ${h}^{-1}\left(x\right)=\text{−}\sqrt{x}$ ( [link] ).

Restricting the domain

Consider the function $f\left(x\right)={\left(x+1\right)}^{2}.$

1. Sketch the graph of $f$ and use the horizontal line test to show that $f$ is not one-to-one.
2. Show that $f$ is one-to-one on the restricted domain $\left[-1,\infty \right).$ Determine the domain and range for the inverse of $f$ on this restricted domain and find a formula for ${f}^{-1}.$
1. The graph of $f$ is the graph of $y={x}^{2}$ shifted left 1 unit. Since there exists a horizontal line intersecting the graph more than once, $f$ is not one-to-one.
2. On the interval $\left[-1,\infty \right),\phantom{\rule{0.2em}{0ex}}f$ is one-to-one.

The domain and range of ${f}^{-1}$ are given by the range and domain of $f,$ respectively. Therefore, the domain of ${f}^{-1}$ is $\left[0,\infty \right)$ and the range of ${f}^{-1}$ is $\left[-1,\infty \right).$ To find a formula for ${f}^{-1},$ solve the equation $y={\left(x+1\right)}^{2}$ for $x.$ If $y={\left(x+1\right)}^{2},$ then $x=-1±\sqrt{y}.$ Since we are restricting the domain to the interval where $x\ge -1,$ we need $±\sqrt{y}\ge 0.$ Therefore, $x=-1+\sqrt{y}.$ Interchanging $x$ and $y,$ we write $y=-1+\sqrt{x}$ and conclude that ${f}^{-1}\left(x\right)=-1+\sqrt{x}.$

Consider $f\left(x\right)=1\text{/}{x}^{2}$ restricted to the domain $\left(\text{−}\infty ,0\right).$ Verify that $f$ is one-to-one on this domain. Determine the domain and range of the inverse of $f$ and find a formula for ${f}^{-1}.$

The domain of ${f}^{-1}$ is $\left(0,\infty \right).$ The range of ${f}^{-1}$ is $\left(\text{−}\infty ,0\right).$ The inverse function is given by the formula ${f}^{-1}\left(x\right)=-1\text{/}\sqrt{x}.$

Inverse trigonometric functions

The six basic trigonometric functions are periodic, and therefore they are not one-to-one. However, if we restrict the domain of a trigonometric function to an interval where it is one-to-one, we can define its inverse. Consider the sine function ( [link] ). The sine function is one-to-one on an infinite number of intervals, but the standard convention is to restrict the domain to the interval $\left[-\frac{\pi }{2},\frac{\pi }{2}\right].$ By doing so, we define the inverse sine function on the domain $\left[-1,1\right]$ such that for any $x$ in the interval $\left[-1,1\right],$ the inverse sine function tells us which angle $\theta$ in the interval $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ satisfies $\text{sin}\phantom{\rule{0.1em}{0ex}}\theta =x.$ Similarly, we can restrict the domains of the other trigonometric functions to define inverse trigonometric functions    , which are functions that tell us which angle in a certain interval has a specified trigonometric value.

Differentiation and integration
yes
Damien
proper definition of derivative
the maximum rate of change of one variable with respect to another variable
terms of an AP is 1/v and the vth term is 1/u show that the sum of uv terms is 1/2(uv+1)
what is calculus?
calculus is math that studies the change in math, such as the rate and distance,
Tamarcus
what are the topics in calculus
Augustine
what is limit of a function?
what is x and how x=9.1 take?
what is f(x)
the function at x
Marc
also known as the y value so I could say y=2x or f(x)= 2x same thing just using functional notation your next question is what is dependent and independent variables. I am Dyslexic but know math and which is which confuses me. but one can vary the x value while y depends on which x you use. also
Marc
up domain and range
Marc
enjoy your work and good luck
Marc
I actually wanted to ask another questions on sets if u dont mind please?
Inembo
I have so many questions on set and I really love dis app I never believed u would reply
Inembo
Hmm go ahead and ask you got me curious too much conversation here
am sorry for disturbing I really want to know math that's why *I want to know the meaning of those symbols in sets* e.g n,U,A', etc pls I want to know it and how to solve its problems
Inembo
and how can i solve a question like dis *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
next questions what do dy mean by (A' n B^c)^c'
Inembo
The sets help you to define the function. The function is like a magic box where you put inside stuff(numbers or sets) and you get out the stuff but in different shapes (forms).
I dont understand what you wanna say by (A' n B^c)^c'
(A' n B (rise to the power of c)) all rise to the power of c
Inembo
Aaaahh
Ok so the set is formed by vectors and not numbers
A vector of length n
But you can make a set out of matrixes as well
I I don't even understand sets I wat to know d meaning of all d symbolsnon sets
Inembo
High-school?
yes
Inembo
am having big problem understanding sets more than other math topics
Inembo
So f:R->R means that the function takes real numbers and provides real numer. For ex. If f(x) =2x this means if you give to your function a real number like 2,it gives you also a real number 2times2=4
pls answer this question *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
If you have f:R^n->R^n you give to your function a vector of length n like (a1,a2,...an) where all a1,.. an are reals and gives you also a vector of length n... I don't know if i answering your question. Otherwise on YouTube you havr many videos where they explain it in a simple way
I would say 24
Offer both
Sorry 20
Actually you have 40 - 4 =36 who offer maths or physics or both.
I know its 20 but how to prove it
Inembo
You have 32+24=56who offer courses
56-36=20 who give both courses... I would say that
solution: In a question involving sets and Venn diagram, the sum of the members of set A + set B - the joint members of both set A and B + the members that are not in sets A or B = the total members of the set. In symbolic form n(A U B) = n(A) + n (B) - n (A and B) + n (A U B)'.
Mckenzie
In the case of sets A and B use the letters m and p to represent the sets and we have: n (M U P) = 40; n (M) = 24; n (P) = 32; n (M and P) = unknown; n (M U P)' = 4
Mckenzie
Now substitute the numerical values for the symbolic representation 40 = 24 + 32 - n(M and P) + 4 Now solve for the unknown using algebra: 40 = 24 + 32+ 4 - n(M and P) 40 = 60 - n(M and P) Add n(M and P), as well, subtract 40 from both sides of the equation to find the answer.
Mckenzie
40 - 40 + n(M and P) = 60 - 40 - n(M and P) + n(M and P) Solution: n(M and P) = 20
Mckenzie
thanks
Inembo
Simpler form: Add the sums of set M, set P and the complement of the union of sets M and P then subtract the number of students from the total.
Mckenzie
n(M and P) = (32 + 24 + 4) - 40 = 60 - 40 = 20
Mckenzie
how do i evaluate integral of x^1/2 In x
first you simplify the given expression, which gives (x^2/2). Then you now integrate the above simplified expression which finally gives( lnx^2).
by using integration product formula
Roha
find derivative f(x)=1/x
-1/x^2, use the chain rule
Andrew
f(x)=x^3-2x
Mul
what is domin in this question
noman
all real numbers . except zero
Roha
please try to guide me how?
Meher
what do u want to ask
Roha
?
Roha
the domain of the function is all real number excluding zero, because the rational function 1/x is a representation of a fractional equation (precisely inverse function). As in elementary mathematics the concept of dividing by zero is nonexistence, so zero will not make the fractional statement
Mckenzie
a function's answer/range should not be in the form of 1/0 and there should be no imaginary no. say square root of any negative no. (-1)^1/2
Roha
domain means everywhere along the x axis. since this function is not discontinuous anywhere along the x axis, then the domain is said to be all values of x.
Andrew
Derivative of a function
Waqar
right andrew ... this function is only discontinuous at 0
Roha
of sorry, I didn't realize he was taking about the function 1/x ...I thought he was referring to the function x^3-2x.
Andrew
yep...it's 1/x...!!!
Roha
true and cannot be apart of the domain that makes up the relation of the graph y = 1/x. The value of the denominator of the rational function can never be zero, because the result of the output value (range value of the graph when x =0) is undefined.
Mckenzie
👍
Roha
Therefore, when x = 0 the image of the rational function does not exist at this domain value, but exist at all other x values (domain) that makes the equation functional, and the graph drawable.
Mckenzie
👍
Roha
Roha are u A Student
Lutf
yes
Roha
What is the first fundermental theory of Calculus?
do u mean fundamental theorem ?
Roha
I want simple integral
for MSc chemistry... simple formulas of integration
aparna
hello?
funny
how are you
funny
I don't understand integration
aparna
r u insane
aparna
integration is so simple not typical..
funny
tell me any questions about integration then i will solve.
funny
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funny
I just formulas
aparna
I just want formulas of integration
aparna
value of log ax cot-x cos-x
aparna
there are many formulas about integration
funny
more then one formula are exist about integration..
funny
so I want simple formulas Because I'm studying MSc chem...Nd have done bsc from bio...
aparna
I am M.sc physics now i am studying in m.phil
funny
so what can i do
aparna
I will send you basic formula for integration after two mint first of all i write then i will send you.
funny
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funny
integration f(X) dx this is basic formula of integration sign is not there you can look integration sign in methematics form... and f(X) my be any function any values
funny
you send me your any ID where i can send you information about integration
funny
funny
Hi
RIZWAN
I don't understand the formula
who's formula
funny
which formula?
Roha
what is the advantages of mathematical economics
Mubarak
What is a independent variable
a variable that does not depend on another.
Andrew
which can be any no... does not need to find its value by any other variable.. often x is independent and y is dependent
Roha
solve number one step by step
x-xcosx/sinsq.3x
Hasnain
x-xcosx/sin^23x
Hasnain
how to prove 1-sinx/cos x= cos x/-1+sin x?
1-sin x/cos x= cos x/-1+sin x
Rochel
how to prove 1-sun x/cos x= cos x / -1+sin x?
Rochel