(2/3)x-1/3 + (2/3)y-1/3 y' = 0, so that (Now solve for y'.) (2/3)y-1/3 y' = - (2/3)x-1/3, and, Since lines tangent to the graph will have slope $ -1 $, set y' = -1, getting, - y 1/3 = -x 1/3, y 1/3 = x 1/3, ( y 1/3) 3 = ( x 1/3) 3, or y = x. Substitue this into the ORIGINAL equation x 2/3 + y 2/3 = 8, getting x 2/3 + (x) 2/3 = 8, 2 x. ISubtitle – the application allows you to create professional subtitles in movies for mobile devices: iPad, iPhone, iPod (Touch, Classic, Nano), Apple TV or for any QuickTime-player. ISubtitle is the first and only application that takes full advantage of Apple's 'soft' subtitle technology. Focus timer 2 8 – focus mind on working capital. First multipling 3.3 we get 9 then again multipling 9 and 2we get 18 so the second part is 18/2 so we get 2+18/2 Taking lcm of denimometer we get 2 and hence going forward we get 4/2+18/2 adding the numeter we get 22 then taking the denimometer common we get 2 so the ans is 22/2 and hence proceeding 22/2=11.
- Isubtitle 3 3 2 Equals What Grade
- Isubtitle 3 3 2 Equals What Mixed Number
- Isubtitle 3 3 2 Equals What Fraction
Purplemath
First you learned (back in grammar school) that you can add, subtract, multiply, and divide numbers. Then you learned that you can add, subtract, multiply, and divide polynomials. Now you will learn that you can also add, subtract, multiply, and divide functions. Performing these operations on functions is no more complicated than the notation itself. For instance, when they give you the formulas for two functions and tell you to find the sum, all they're telling you to do is add the two formulas. There's nothing more to this topic than that, other than perhaps some simplification of the expressions involved.
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Given f (x) = 3x + 2 and g(x) = 4 – 5x, find (f + g)(x), (f – g)(x), (f × g)(x), and (f / g)(x).
To find the answers, all I have to do is apply the operations (plus, minus, times, and divide) that they tell me to, in the order that they tell me to.
(f + g)(x) = f (x) + g(x)
= [3x + 2] + [4 – 5x]
= 3x + 2 + 4 – 5x
= 3x – 5x + 2 + 4
= –2x + 6
(f – g)(x) = f (x) – g(x)
= [3x + 2] – [4 – 5x]
= 3x + 2 – 4 + 5x
= 3x + 5x + 2 – 4
= 8x – 2
(f × g)(x) = [f (x)][g(x)]
= (3x + 2)(4 – 5x)
= 12x + 8 – 15x2 – 10x
= –15x2 + 2x + 8
My answer is the neat listing of each of my results, clearly labelled as to which is which.
( f + g ) (x) = –2x + 6
( f – g ) (x) = 8x – 2
( f × g ) (x) = –15x2 + 2x + 8
(f /g)(x) = (3x + 2)/(4 – 5x)
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Given f (x) = 2x, g(x) = x + 4, and h(x) = 5 – x3, find (f + g)(2), (h – g)(2), (f × h)(2), and (h / g)(2).
This exercise differs from the previous one in that I not only have to do the operations with the functions, but I also have to evaluate at a particular x-value. To find the answers, I can either work symbolically (like in the previous example) and then evaluate, or else I can find the values of the functions at x = 2 and then work from there. It's probably simpler in this case to evaluate first, so:
f (2) = 2(2) = 4
g(2) = (2) + 4 = 6
h(2) = 5 – (2)3 = 5 – 8 = –3
Now I can evaluate the listed expressions:
(f + g)(2) = f (2) + g(2)
(h – g)(2) = h(2) – g(2)
= –3 – 6 = –9
(f × h)(2) = f (2) × h(2)
(h / g)(2) = h(2) ÷ g(2)
= –3 ÷ 6 = –0.5
Then my answer is:
(f + g)(2) = 10, (h – g)(2) = –9, (f × h)(2) = –12, (h / g)(2) = –0.5
If you work symbolically first, and plug in the x-value only at the end, you'll still get the same results. Either way will work. Evaluating first is usually easier, but the choice is up to you.
You can use the Mathway widget below to practice operations on functions. Try the entered exercise, or type in your own exercise. Then click the button and select 'Solve' to compare your answer to Mathway's. (Or skip the widget and continue with the lesson.)
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(Clicking on 'Tap to view steps' on the widget's answer screen will take you to the Mathway site for a paid upgrade.)
Givenf (x) = 3x2 – x + 4, find the simplified form of the following expression, and evaluate at h = 0:
This isn't really a functions-operations question, but something like this often arises in the functions-operations context. This looks much worse than it is, as long as I'm willing to take the time and be careful.
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Isubtitle 3 3 2 Equals What Grade
The simplest way for me to proceed with this exercise is to work in pieces, simplifying as I go; then I'll put everything together and simplify at the end.
For the first part of the numerator, I need to plug the expression 'x + h' in for every 'x' in the formula for the function, using what I've learned about function notation, and then simplify:
f(x + h)
= 3(x + h)2 – (x + h) + 4
Isubtitle 3 3 2 Equals What Mixed Number
= 3(x2 + 2xh + h2) – x – h + 4
= 3x2 + 6xh + 3h2 – x – h + 4
The expression for the second part of the numerator is just the function itself:
Now I'll subtract and simplify:
f(x + h) – f(x)
= [3x2 + 6xh + 3h2 – x – h + 4] – [3x2 – x + 4]
= 3x2 + 6xh + 3h2 – x – h + 4 – 3x2 + x – 4
= 3x2 – 3x2 + 6xh + 3h2 – x + x – h + 4 – 4
= 6xh + 3h2 – h
All that remains is to divide by the denominator; factoring lets me simplify:
Now I'm supposed to evaluate at h = 0, so:
6x + 3(0) – 1 = 6x – 1
simplified form: 6x + 3h – 1
value at h = 0: 6x – 1
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That's pretty much all there is to 'operations on functions' until you get to function composition. Don't let the notation for this topic worry you; it means nothing more than exactly what it says: add, subtract, multiply, or divide; then simplify and evaluate as necessary. Don't overthink this. It really is this simple.
Oh, and that last example? They put that in there so you can 'practice' stuff you'll be doing in calculus. You likely won't remember this by the time you actually get to calculus, but you'll follow a very similar process for finding something called 'derivatives'.
URL: https://www.purplemath.com/modules/fcnops.htm
Purplemath
You've been playing with 'y=' sorts of equations for some time now. And you've seen that the 'nice' equations (straight lines, say, rather than ellipses) are the ones that you can solve for 'y=' and then plug into your graphing calculator. These 'y=' equations are functions. But the question you are facing at the moment is 'Why do I need this function notation, especially when I've got perfectly nice 'y=', and how does this notation work?'
Think back to when you were in elementary school. Your teacher gave you worksheets containing statements like '[ ] + 2 = 4' and told you to fill in the box. Once you got older, your teacher started giving you worksheets containing statements like 'x + 2 = 4' and told you to 'solve for x'.
MathHelp.com
Why did your teachers switch from boxes to variables? Well, think about it: How many shapes would you have to use for formulas like the one for the area A of a trapezoid with upper base a, lower base b, and height h? The formula is as follows:
A = (h/2 )(a + b)
If you try to express the above, or something more complicated, using variously-shaped boxes, you'd quickly run out of shapes. Besides, you know from experience that 'A' stands for 'area', 'h' stands for 'height', and 'a' and 'b' stand for the lengths of the parallel top and bottom sides. Heaven only knows what a square box or a triangular box might stand for!
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In other words, they switched from boxes to variables because, while the boxes and the letters mean the exact same thing (namely, a slot waiting to be filled with a value), variables are better. Variables are more flexible, easier to read, and can give you more information.
The same is true of 'y' and 'f (x)' (pronounced as 'eff-of-eks'). For functions, the two notations mean the exact same thing, but 'f (x)' gives you more flexibility and more information. You used to say 'y = 2x + 3; solve for y when x = –1'. Now you say 'f (x) = 2x + 3; find f (–1)' (pronounced as 'f-of-x equals 2x plus three; find f-of-negative-one'). In either notation, you do exactly the same thing: you plug –1 in for x, multiply by the 2, and then add in the 3, simplifying to get a final value of +1.
But function notation gives you greater flexibility than using just 'y' for every formula. For instance, your graphing calculator will list different functions as y1, y2, etc, so you can tell the equations apart when, say, you're looking at their values in 'TABLE'.
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In the same way, in textbooks and when writing things out, we use different function names like f (x), g(x), h(x), s(t), etc, to keep track of, and work with, more than one formula in any single context. With function notation, we can now use more than one function at a time without confusing ourselves or mixing up the formulas, leaving ourselves wondering 'Okay, which 'y' is this one?' And the notation can be usefully explanatory.
From geometry, you know that 'A(r) = πr2' indicates the area of a circle given in terms of the value of the radius r, while 'C(r) = 2πr' indicates the circumference given in terms of the radius r. Both functions have the same plug-in variable (the 'r'), but 'A' reminds you that the first function is the formula for 'area' and 'C' reminds you that the second function is the formula for 'circumference'.
Remember: The notation 'f (x)' is exactly the same thing as 'y'. You can even label the y-axis on your graphs with 'f (x)', if you feel like it.
Let me clarify another point. While parentheses have, up until now, always indicated multiplication, that is not the case with function notation. Contrary to all previous experience, the parentheses for function notation do not indicate multiplication.
Isubtitle 3 3 2 Equals What Fraction
The expression 'f (x)' means 'a formula, named f, has x as its input variable'. It does not mean 'multiply f and x'!
Don't embarrass yourself by pronouncing (or thinking of) 'f (x)' as being 'f times x', and never try to 'multiply' the function name with its parenthesised input.
In function notation, the 'x' in 'f (x)' is called 'the argument of the function', or just 'the argument'. So if they give you the expression 'f (2)' and ask for the 'argument', the answer is just '2'.
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Aside: Why is the input called the 'argument'? The term 'argument' has a long history. Originally, it was a logical term, referring to a statement that forwarded a proof or, in a less formal sense, a claim that was being used to try to convince somebody of something. Eventually, the term came to refer, in early scientific contexts, to any mathematical value that was needed as an input to other computations, or any value upon which later results depended.
In the twentieth century, when computer coding started becoming a thing, coders adopted the mathematical meaning to refer to inputs to their coding. In our mathematical context, the 'argument' is the independent variable (the one for which you pick a value, usually being the x-value) and the function's output is the dependent variable (the one whose value depends upon whatever was plugged in, usually being the y-value).
Given h(s), what is the function name, and what is the argument?
I'll do the second part first. The argument is whatever is inside the parentheses, so the argument here is s.
The function name is the variable that comes before the parentheses. In this case, then, the function name is h.
What is the argument of f (y)?
The argument is whatever is plugged in. In this particular (unusual) case, the variable being plugged in is 'y'. (After all, there's no rule saying that y can't be the independent variable.) So:
Given g(t) = t 2 + t, what is the function name? In g(–1), what is the argument?
The function name is what comes before the parentheses, so the function name here is g.
In the second part of the question, they're asking me for the argument. In the first part, where they gave me the function name and argument (being the 'g(t)' part) and the formula (being the 't 2 + t' part), the argument was t. But in the second part, they've plugged a particular value in for t. So, in the second part, the argument is the number –1.
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Evaluating at a Number
You evaluate 'f (x)' in exactly the same way that you've always evaluated 'y'. Namely, you take the number they give you for the input variable, you plug it in for the variable, and you simplify to get the answer. For instance:
Given f (x) = x2 + 2x – 1, find f (2).
To evaluate f (x) at x = 2, I'll plug 2 in for every instance of x in the function's rule:
To keep things straight in my head (and clear in my working), I've put parentheses around every instance of the argument 2 in the formula for f. Now I can simplify: Balsamiq mockups.
(2)2 + 2(2) – 1
Then my answer is:
f (2) = 7
Given f (x) = x2 + 2x – 1, find f (–3).
To evaluate, I do what I've always done. I'll plug the given value (–3) in for the specified variable (x) in the given formula:
Once again, I've used parentheses to clearly designate the value being input into the formula. In this case, the parentheses are helping me keep track of the 'minus' signs. Now I can simplify:
(–3)2 + 2(–3) – 1
Then my answer is:
f (–3) = 2
If you experience difficulties when working with negatives, try using parentheses as I did above. Doing so helps keep track of things like whether or not the exponent is on the 'minus' sign. And it's just generally a good habit to develop.
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An important type of function is called a 'piecewise' function, so called because, well, it's in pieces. For instance, the following is a piecewise function:
= 1'>
As you can see, this function is split into two halves: the half that comes before x = 1, and the half that goes from x = 1 to infinity. Which half of the function you use depends on what the value of x is. Let's examine this:
Given the function f (x) as defined above, evaluate the function at the following values: x = –1, x = 3, and x = 1.
This function comes in pieces; hence, the name 'piecewise' function. When I evaluate it at various x-values, I have to be careful to plug the argument into the correct piece of the function.
They first want me to evaluate at x = –1. Since this is less than 1, then this argument goes into the first piece of the function. To refresh, the function is this:
Then I'll be plugging the –1 into the rule 2x2 – 1:
f (–1) = 2(–1)2 – 1
Next, they want me to find the value of f (3). Since 3 is greater than 1, then I'll need to plug into the second piece of the function, so:
f (3) = (3) + 4 = 7
Finally, they want me to evaluate f (x) at x = 1. This is the only x-value that's a little tricky. Which half do I use?
Looking carefully at the rules for the functions, I can see that the first piece is the rule for x-values that are strictly less than 1; the rule does not apply when x equals 1. On the other hand, the second piece applies when x is greater than or equal to1. Since I'm dealing here with x = 1, then the second piece's rule applies.
Then my answer is:
f (–1) = 1
f (3) = 7
f (1) = 5
You can use the Mathway widget below to practice evaluating functions at a given numerical value. Try the entered exercise, or type in your own exercise. Then click the button and select 'Evaluate' to compare your answer to Mathway's.
Please accept 'preferences' cookies in order to enable this widget.
(Click 'Tap to view steps' to be taken directly to the Mathway site for a paid upgrade.)
URL: https://www.purplemath.com/modules/fcnnot.htm
