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  1. "x += 1" vs "x = x + 1" : r/learnpython - Reddit

    Apr 16, 2020 · I honestly don't understand how this expression can be mathematically correct. I always end up writing x=+1 by accident because the correct way of writing seems so …

  2. algebra precalculus - Factorize $ (x+1) (x+2) (x+3) (x+6)- 3x^2 ...

    Aug 9, 2013 · Factorize : $$ (x+1) (x+2) (x+3) (x+6)- 3x^2$$ I don't really know where to start, but I expanded everything to get : $$x^4 + 12x^3 + 44x^2 + 72x + 36$$ I used rational roots test …

  3. algebra precalculus - Factoring $ (x+1) (x+2) (x+3) (x+4)+1 ...

    Aug 25, 2014 · If you can pretend you "guessed" the factorization, notice $ (x^2+5x+5)^2-1$ is a difference of squares.

  4. Prove $ x^n-1=(x-1)(x^{n-1}+x^{n-2}+\\dots+x+1)$

    Aug 17, 2014 · My base case is when n= 2 n = 2 we have on the left side of the equation x2 −1 x 2 1 and on the right side: (x−1)(x+1) (x 1) (x + 1) which when distributed is x2 −1 x 2 1.

  5. calculus - How to determine if a function is one-to-one?

    I am looking for the "best" way to determine whether a function is one-to-one, either algebraically or with calculus. I know a common, yet arguably unreliable method for determining this answer …

  6. Derivative of $e^ {x+1}$ - Mathematics Stack Exchange

    Aug 16, 2016 · How did you get your answer? Did you try to apply the chain rule? You might get better answers if you show your process so we can point out where things go wrong.

  7. Verify $(y-x)y' = y-x+8$ has an explicit solution $y = x+ 4 …

    Jan 26, 2017 · Ive solved this problem and both sides of the function don't appear to be equal to me. However the solution in the back of the book says it is a solution. Am I not simplifying …

  8. Solutions of the functional equation $f (x+1)= xf (x)$

    Mar 21, 2021 · $$g (x+1)=g (x)+\log x.$$ The solution of such recurrences is made of the general solution of the homogeneous equation (obviously a constant), and a particular solution.

  9. Prove inequality using mean value theorem 1/(x+1) < ln(x+1)

    Feb 7, 2022 · Closed 3 years ago. Using the mean value theorem of lagrange I need to prove that for all x > 0: $$ \frac {1} {x+1} < ln (x+1) - ln (x) < \frac {1} {x} $$

  10. Does $x = x+1$ have infinitely many solutions? [closed]

    Mar 1, 2023 · If $x = x+1,$ then, by substitution, we have $x+1 = x+1.$ Does this mean that $x = x+1$ has infinitely many solutions?