Log in Monika 12 years agoPosted 12 years ago. Direct link to Monika's post “Could you simplify it fur...” Could you simplify it further to: x(7x^3-5x^2+28x-20) ? • (37 votes) Nicholas 12 years agoPosted 12 years ago. Direct link to Nicholas's post “You aren't making it any ...” You aren't making it any simpler by factoring out an x. If anything it complicates it. (100 votes) Emma 11 years agoPosted 11 years ago. Direct link to Emma's post “At 1:09 Sal says that he ...” At • (10 votes) ArDeeJ 11 years agoPosted 11 years ago. Direct link to ArDeeJ's post “FOIL won't help you if yo...” FOIL won't help you if you have to expand a product that isn't two binomials multiplied together; for example, two trinomials multiplied together. It's usually better to understand what you're doing instead of relying on mnemonics. For example: (a + b + c) * (d + e + f) (30 votes) Diego Perez 11 years agoPosted 11 years ago. Direct link to Diego Perez's post “At 2:28 why is it that 7x...” At • (3 votes) xanderspeer 7 years agoPosted 7 years ago. Direct link to xanderspeer's post “but 7x means X+X+X+X+X+X+...” but 7x means X+X+X+X+X+X+X (1 vote) Jamarcus Hull 11 years agoPosted 11 years ago. Direct link to Jamarcus Hull's post “How do you find the produ...” How do you find the product of this question? (x+1)(-5x+7) • (0 votes) Kael Ruland 11 years agoPosted 11 years ago. Direct link to Kael Ruland's post “Alright, so we use someth...” Alright, so we use something called FOIL, which stands for "firsts, outers, inners, lasts". When we multiply binomials like this, we sum the products of each of the components of FOIL. (a+b)(c+d) = ac+ad+bc+bd (23 votes) Melenie Mattingly 11 years agoPosted 11 years ago. Direct link to Melenie Mattingly's post “what do you mean by (fx) ...” what do you mean by (fx) =7x -5 is this a formula • (4 votes) Petrie (Peter S. Asiain III) 11 years agoPosted 11 years ago. Direct link to Petrie (Peter S. Asiain III)'s post “This function f(`x`) =7`...” This function f( (7 votes) Saleh Hussain 5 years agoPosted 5 years ago. Direct link to Saleh Hussain's post “In 2:14 Why x^3 * 7x = 7...” In • (3 votes) kubleeka 5 years agoPosted 5 years ago. Direct link to kubleeka's post “We have x³·7x. That is, x...” We have x³·7x. That is, x·x·x·7·x. We can rearrange it into 7·x·x·x·x, which is 7x⁴. (4 votes) Petra 10 years agoPosted 10 years ago. Direct link to Petra's post “What do you do if you're ...” What do you do if you're given a number? For example, you're given your two equations and it says: Find (f o g) (0)? Do you plug the zero into your equations or solve f*g then input the 0 into your output? • (4 votes) abhi.devata 8 years agoPosted 8 years ago. Direct link to abhi.devata's post “You could do either way....” You could do either way. (2 votes) Chaya Zajac 11 years agoPosted 11 years ago. Direct link to Chaya Zajac's post “When we got the final fun...” When we got the final function after solving the problem, Sal stated that we cannot simplify the expression because each variable was to a different degree. But, I thought one can subtract variables with exponents, so why can one not simplify the expression? • (3 votes) Chaya Zajac 11 years agoPosted 11 years ago. Direct link to Chaya Zajac's post “What happens when we have...” What happens when we have a^2+a^3? Can we simplify that? (3 votes) Aimee 12 years agoPosted 12 years ago. Direct link to Aimee's post “is (f*g)(x) the same as (...” is (f*g)(x) the same as (f o g)(x)? • (1 vote) karan1276 12 years agoPosted 12 years ago. Direct link to karan1276's post “no, (f o g)(x) = f(g(x))...” no, (f o g)(x) = f(g(x)) (7 votes) M93A.cz 10 years agoPosted 10 years ago. Direct link to M93A.cz's post “Hi. I'm primary a program...” Hi. I'm primary a programmer but I'm really interested in math and I think I found another possible interpretation of multiplying functions. It gives a clue that there could possibly be NOTE: In latex The Is this used somewhere in the mathematics? • (3 votes) ArDeeJ 10 years agoPosted 10 years ago. Direct link to ArDeeJ's post “Yes, that's all correct. ...” Yes, that's all correct. f^2(x) is commonly taken to mean f(f(x)), similarly with higher exponents. Negative exponents refer to the inverse function, that is, f^(-2)(x) = f^(-1)(f^(-1)(x)). What you called \times is called function composition, and is written (g ∘ f)(x) = g(f(x)). As you noted, it's not commutative, but it is associative. Whenever the compositions are defined, (h ∘ g) ∘ f = h ∘ (g ∘ f) = h ∘ g ∘ f. In a way, the function iteration can be extended to fractional exponents as well. For example, the function g(x) = f^(1/2)(x) would be a function that satisfies g^2(x) = f(x). Also, as a side note, the neutral function is more commonly called the identity function (and the neutral element 1 is called the identity element). (The trigonometric functions break this convention: sin^2(x) is taken to mean sin(x)*sin(x). However, sin^(-1)(x) = arcsin x still refers to the inverse function. Arcsin is preferred over sin^(-1).) (3 votes)Want to join the conversation?
But yes you could if you wanted to.
1:09
= ad + ae + af + bd + be + bf + cd + ce + cf 2:28
so if you multiply it by an exponent shouldnt you get a different answer?x) =7x−5 means that each time we plug in a value of x we would multiply it by 7 then subtract 5.
f(x)=7x−5 → This our function with just x.
Lets try substituting different values for x
f(3)=73−5=21−5=16 → If substitute x by 3, this is what we get.
f(8)=78−5=56−5=51 → If substitute x by 8, this is what we get. 2:14
Why x^3 * 7x = 7x^4 !
Why there's fourth power. I can't get it :(
Personally, I prefer to solve both equations for the given number first, then combine them. But it doesn't really matter
but (f*g)(x) = f(x)*g(x)
i think thats what it is ,but i am not SURE!!
The key is in the notation f^-1 which is defined as
where \forall functions f and x \in D(f) and y \in R(f):
y=f(x) <=> x=(f^-1)(y)D(f) is the domain and R(f) is the range of a function.f^2 meaning f(f(x)).
This would mean that (f \times g)(x) = f(g(x)).\times is a vector cross product. I use to distinguish it from the ordinary multiplication. The same as cross product, this type of multiplication is also not commutative.\times operator treats similar to ordinary multiplication of real numbers:\forall x \in R-{0}: x * x^-1 = 1 where 1 is the neutral element of multiplication.\forall functions f: f \times f^-1 = n where n is the neutral function n(x)=x.
