1. Solve for x: \(\frac{2x^2+x+3}{x^2+x+1}=\frac{2x+1}{x+1}\)
2. Let g(x) be a function piecewise defined as \(g(x) = \left\{ \begin{array}{cl} -x & x\le 0, \\ 2x-41 & x>0. \end{array} \right.\). If a is negative, find a so that g(g(g(10.5)))=g(g(g(a))).
This is for the first problem.
We can start by cross-multiplying the two expressions:
(2x^2 + x + 3)(x + 1) = (2x + 1)(x^2 + x + 1)
Expanding both sides, we get:
2x^3 + 3x^2 + 3x + 2x^2 + 2x + 3 = 2x^3 + 3x^2 + 2x + x^2 + x + 1
Combining like terms, we get:
5x^2 + 5x + 4 = 3x^3 + 4x^2 + 3x + 1
Moving all the terms to the left-hand side, we get:
3x^3 - 5x^2 - 2x - 3 = 0
We can factor this expression as:
(x - 1)(3x^2 + 4x + 3) = 0
The first factor, x−1, will always be nonzero, so the only solution is x=1.
+129771 1.
We can start by cross-multiplying the two expressions:
(2x^2 + x + 3)(x + 1) = (2x + 1)(x^2 + x + 1)
2x^3 + x^2 + 3x + 2x^2 + x + 3 = 2x^3 + 2x^2 + 2x + x^2 + x + 1
2x^3 + 3x^2 + 4x + 3 = 2x^3 + 3x^2 + 3x + 1
4x + 3 = 3x + 1
x = 1 - 3
x = -2
This is for problem 2!
First, we evaluate g(10.5). Since 10.5 is positive, we use the second definition of g(x), so g(10.5) = 2(10.5) - 41 = -20.5.
Next, we evaluate g(-20.5). Since -20.5 is negative, we use the first definition of g(x), so g(-20.5) = -(-20.5) = 20.5.
Finally, we evaluate g(20.5). Since 20.5 is positive, we use the second definition of g(x), so g(20.5) = 2(20.5) - 41 = -6.
We are given that g(g(g(10.5)))=g(g(g(a))). Since g(10.5) = -20.5 and g(-20.5) = 20.5, we have g(g(g(10.5))) = g(g(20.5)). Since g(20.5) = -6, we have g(g(20.5)) = g(-6). Since g(-6) = 6, we have g(g(g(10.5))) = 6.
We are also given that a is negative. Since g(g(g(a))) = 6, we have g(a) = -6. Since g(a) = -6 and a is negative, we have a = -6.
Therefore, the value of a is −6.
