How do I simplify "x/(x-1)". I know it simplifies to: "1+1/(x-1)," but how?
x / (x - 1) = (x - 1 + 1) / (x - 1) . . . . . . . . = [(x - 1) + 1] / (x - 1) . . . . . . . . = [(x - 1)/(x - 1)] + [1 / (x - 1)] . . . . . . . . = 1 + [1 / (x - 1)] Alejandra
Simplify fully: 2/(x-1) + x-11/(x^2+3x-4)?
2/(x-1) + x-11/(x^2+3x-4) Factor the trinomial 2/(x-1) + x-11/(x-1)(x+4) make the denominators the same by multiplying (x+4) on the left 2(x+4)/(x-1)(x+4) + x-11/(x-1)(x+4) distribute the 2 2x+8+x-11/(x-1)(x+4) combine like terms 3x-3/(x-1)(x+4) remove the 3 from the top equation 3(x-1)/(x-1)(x+4) *note: 3(x-1) is the same as 3x-3 cancel (x-1) on the top and bottom 3/x+4 <---answer
How do I simplify f(x) =2*f(x-1)-1?
f(x) = 2*f(x-1)-1f(x) = 2*(2*f(x-2)-1)-1Inductionf(x) = (2^n)*f(x-n)-1–2–4…-2^(n-1)f(x) = (2^n)*f(x-n)-(2^n-1)n approaches xf(x) = 2^x*f(0)-2^x+1f(x) = 2^x(f(0)-1)+1I still need an initial value, for f(0) to give an answer that totally depends on x, though.
Can you simplify x-1 = x+1 ?
Yes. Subtract X-1 from both sides to get(x-1) - (x-1) = (x+1) - (x-1) this gives you0 = 2 ??? OOPS something is wrong! Lets look at this another way:Let x = 2 Then, form the equation,x² = 2x True! Then, subtract 4 from both sides:x² - 4 = 2x - 4 True! Now, factor each side without changing any values:(x + 2) (x - 2) = 2 (x - 2) Done properly! Now, divide by both sides by (x - 2) to get(x + 2) = 2 Simplified! Therefore,x = 2 - 2 = 0 !!!!.!.!….!……..???? didn’t we start with x = 2??? we must then be saying:x = 2 = 0 So therefore,2 = 0 !!!!……????? what went wrong??? Or did we just prove that2 always equals 0?This tells me that I can eat 2 hamburgers at every meal and still lose weight! (Oh, how I wish!!! but probably that won’t happen!)Why does this not work? All of the algebra seems to be in order. For the answer, look at the line where we divided both sides by the term (x - 2). Since we had originally defined x = 2, then the term we divided by can be evaluated this way…IF x = 2, then x - 2 = 2 - 2 = 0Thus we divided both sides by zero, a process that is undefined. Although, lets try sneaking up on the answer: The limit of (x - 2) as x approaches 2 is zero. And any non-zero number divided by the term (x - 2) as x approaches 2 is infinite, also undefined. Thus all steps beyond this one are invalid. Although…I…..I am still wishing I could eat two hamburgers at every meal and still lose weight!QED
How do I factor completely and simplify 2x^2+4x+2 answer and steps please?
First take out the GCF (Greatest Common Factor). You always look first to see if there is a GCF. Here it would be 2. 2x^2+4x+2 --> 2 (x^2+2x+1) Ignore the 2 that was factored out for now. Now there is a polynomial.You have to find 2 numbers that multiply to the last term and add to the second term. The last term is 1 and the second term is 2 (ignore the variable). Luckily, only 1 x 1 multiply to 1 . So the two number that we have to use is 1 and 1. And 1 +1 adds to 2 obviously. In order to get a trinomial, you always have to multiply two binomials. So when we are factoring a trinomial we set up two binomials, because factoring is essentially going backwards. ( ) ( ) In order to get the first term of the polynomial ( x^2) we would have to multiply x times x. So, (x ) (x ). Now we know that our second term has to be 1and 1 because those are the numbers that multiply to 1 and add to 2. So, (x 1 ) (x 1). Here The signs are going to be positive because in the trinomial there is a positive 2x and a positive 1. So, (x+1 )(X+1) Now we bring back the 2 we factored out earlier, and we place it in front of the 2 binomials. 2 (x+1)(x+1) You can check your answer but multiplying this whole thing out. When you do, you should get the problem you first started with.
For real x != 0 let f(x) = xcos(1/x) and define f(0) = 0.?
first you should know the left hand derivative at x = 0 and right hand derivative at x = 0 and show both are finite and both are equal to each other LHD = lim x -> 0- [f(x) - f(0)]/ (x-0) this limit means x approaches 0 from left hand side of the number line through negative values. RHD = lim x ->0+ [f(x) -f(0)]/(x-0) this is limit when x approaches 0 from positive side this limit exists only if both limits are equal now f(0) is defined as 0 we need to see if both these limits exist limit x->0 f(x)/x = xcos(1/x) /x = cos(1/x) which is the same as limit y->infinity cosy where y = 1/x but you say this limit does not exist hence this is not differentiable at x = 0 hope this is clear.
How do you expand and simplify (x+2) (x-3)?
In multiplying the two given binomials, x + 2 and x - 3, we can use the "FOIL" method and then simply as follows:In using the "FOIL" method, we'll begin by multiplying the two first (F) terms of the two binomials, i.e., x times x, then, multiply together the two outside (O) terms, i.e., x and -3, then, multiply together the two inside (I) terms, i.e., 2 and x, and finally, we'll multiply together the two last (L) terms, i.e., 2 and -3, and then simplify these partial products:(x + 2)(x - 3) = (x + 2)(x + (-3)) = x(x) + (-3)(x) + 2(x) + 2(-3) = x^2 + (-3x) + 2x + (- 6) = x^2 + (-3x) + 2x - 6 Now, collecting like-terms, we have: = x^2 + [(-3) + 2]x - 6 = x^2 + [-1]x - 6 = x^2 - x - 6 is the trinomial that is the final result after simplifying.
Is x-1/x+1 a polynomial?
This is a rational function,a polynomial is if the form ax^n+bx^(n-1)+cx^(n-3)+.........
How do you factorize x³-x²-x+1?
First, notice that replacing all the x's with 1's gives you 1-1-1+1, which equals 0. This is the first thing you should check with a polynomial that you wish to factor. Since 1 is a root of x³-x²-x+1=0, (x-1) is a factor of x³-x²-x+1. Divide x³-x²-x+1 by x-1 to get x²-1. (You can check this with long division if it's not immediately apparent). x²-1 is further factored into x-1 and x+1. So the factors of x³-x²-x+1 are x-1, x-1, and x+1.
How do I factorize x^2 - 1-2y-y^2?
x^2–1–2y-y^2=(x)^2-(1+2y+y^2)=(x)^2 - (y+1)^2=(x+y+1) (x-y-1) , Answer.