In many areas of science, engineering, and mathematics, it is useful to know the maximum value a function can obtain, even if we don’t know its exact value at a given instant. For instance, if we have a function describing the strength of a roof beam, we would want to know the maximum weight the beam can support without breaking. If we have a function that describes the speed of a train, we would want to know its maximum speed before it jumps off the rails. Safe design often depends on knowing maximum values.
This project describes a simple example of a function with a maximum value that depends on two equation coefficients. We will see that maximum values can depend on several factors other than the independent variable
x .
Consider the graph in
[link] of the function
$y=\text{sin}\phantom{\rule{0.1em}{0ex}}x+\text{cos}\phantom{\rule{0.1em}{0ex}}x.$ Describe its overall shape. Is it periodic? How do you know?
Using a graphing calculator or other graphing device, estimate the
$x$ - and
$y$ -values of the maximum point for the graph (the first such point where
x >0). It may be helpful to express the
$x$ -value as a multiple of π.
Now consider other graphs of the form
$y=A\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}x+B\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}x$ for various values of
A and
B . Sketch the graph when
A = 2 and
B = 1, and find the
$x$ - and
y -values for the maximum point. (Remember to express the
x -value as a multiple of π, if possible.) Has it moved?
Repeat for
A = 1,
B = 2. Is there any relationship to what you found in part (2)?
Complete the following table, adding a few choices of your own for
A and
B :
A
B
x
y
A
B
x
y
0
1
$\sqrt{3}$
1
1
0
1
$\sqrt{3}$
1
1
12
5
1
2
5
12
2
1
2
2
3
4
4
3
Try to figure out the formula for the
y -values.
The formula for the
$x$ -values is a little harder. The most helpful points from the table are
$\left(1,1\right),\left(1,\sqrt{3}\right),\left(\sqrt{3},1\right).$ (
Hint : Consider inverse trigonometric functions.)
If you found formulas for parts (5) and (6), show that they work together. That is, substitute the
$x$ -value formula you found into
$y=A\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}x+B\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}x$ and simplify it to arrive at the
$y$ -value formula you found.
Key concepts
For a function to have an inverse, the function must be one-to-one. Given the graph of a function, we can determine whether the function is one-to-one by using the horizontal line test.
If a function is not one-to-one, we can restrict the domain to a smaller domain where the function is one-to-one and then define the inverse of the function on the smaller domain.
For a function
$f$ and its inverse
${f}^{\mathrm{-1}},f\left({f}^{\mathrm{-1}}\left(x\right)\right)=x$ for all
$x$ in the domain of
${f}^{\mathrm{-1}}$ and
${f}^{\mathrm{-1}}\left(f\left(x\right)\right)=x$ for all
$x$ in the domain of
$f.$
Since the trigonometric functions are periodic, we need to restrict their domains to define the inverse trigonometric functions.
The graph of a function
$f$ and its inverse
${f}^{\mathrm{-1}}$ are symmetric about the line
$y=x.$
For the following exercises, a. find the inverse function, and b. find the domain and range of the inverse function.
$f\left(x\right)={x}^{2}-4,x\ge 0$
a.
${f}^{\mathrm{-1}}\left(x\right)=\sqrt{x+4}$ b. Domain
$\text{:}\phantom{\rule{0.2em}{0ex}}x\ge \mathrm{-4},\text{range}\text{:}\phantom{\rule{0.2em}{0ex}}y\ge 0$
(mathematics) For a complex number a+bi, the principal square root of the sum of the squares of its real and imaginary parts, √a2+b2 . Denoted by | |.
The absolute value |x| of a real number x is √x2 , which is equal to x if x is non-negative, and −x if x is negative.
Hello please can someone tell me the meaning of this group all about, yes I know is calculus group but yet nothing is showing up
Shodipo
You have downloaded the aplication Calculus Volume 1, tackling about lessons for (mostly) college freshmen, Calculus 1: Differential, and this group I think aims to let concerns and questions from students who want to clarify something about the subject.
Well, this is what I guess so.