Understanding Algebra for Calculus

1. The function \(f\) tells us to take any input \(x\) and do the following:

   S1:

   square the input \(x\)

   S2:

   multiply S1 by \(3\)

   S3:

   subtract \(1\) from S2.

   If \(x=2\text{,}\) we will get: \(2^2 = 4\text{;}\) then multiply by \(3\) to get \(12\text{;}\) then subtract \(1\) to get \(11\text{.}\) So \(f(2) = 3\cdot 2^2 -1 = 11\text{.}\) The output of \(f\) when \(x=2\) is \(y = 11\text{.}\)

2. This time, we think of \(x+1\) as our input, a quantity that must be evaluated before \(f\) is applied, so we enclose it in parentheses in the evaluation. Then we apply the steps and simplify by expanding and collecting like terms.

   \begin{align*} f(x+1) \amp = \amp 3(x+1)^2 - 1\\ \amp = \amp 3(x^2+2x+1) - 1 \amp \text{ square } (x+1) \\ \amp = \amp 3x^2+6x +3 - 1 \amp \text{ multiply by } 3 \\ \amp = \amp 3x^2+6x + 2. \amp \text{ subtract } 1 \end{align*}

   Note that \(y=f(x+1)\) does not define the same function as \(y=f(x)\text{;}\) for any \(x\)-value, \(y=f(x+1)\) says that we add \(1\) to \(x\) and then apply all the steps of \(f\text{.}\) This can be used to define another function, say \(y = g(x) = f(x+1)\text{.}\)

3. In this part we first apply \(f\) to \(x\text{,}\) and then do two additional steps: multiply by \(2\) and then subtract \(1\text{.}\)

   S4:

   multiply \(f(x)\) by \(2.\)

   S5:

   subtract \(1\) from S4.

   Here we think of \(f(x)\) as being enclosed in parentheses and simplify:

   \begin{align*} 2f(x) - 1 \amp = \amp 2(3x^2-1) -1 \\ \amp = \amp 6x^2-2-1 \\ \amp = \amp 6x^2-3. \end{align*}

S1:

First we subtract \(2\) from \(x\text{.}\) There is no value of \(x\) that cannot have \(2\) subtracted from it, so we have no restictions yet.

S2:

Now take the square root of of the result of S1. We cannot put a negative value under the square root sign; thus \(x-2\) (the result of S1 must be nonnegative. Set \(x-2 \geq 0\text{,}\) we can solve this to find \(x \geq 2\text{.}\) Thus the domain is \([2, \infty)\text{.}\)

Numerator

S1:

Square \(x\text{.}\)

S2:

Subtract \(1\) from S1.

S3:

Take the square root of S2.

Denominator

S4:

Multiply \(x\) by 2.

S5:

Add \(3\) to S4.

Divide Numerator by Denominator

We can look at each part separately to determine the domain. For Numerator, we can certainly square any \(x\text{,}\) and subtract \(1\text{.}\) It is the square root that will be a problem. We need to have \(x^2-1 \geq 0\text{.}\) If \(x\) is between \(-1\) and \(1\text{,}\) then \(x^2\) is less than \(1\text{,}\) so \(x^2-1\) is negative. So \(-1<x<1\) must be excluded from the domain since these values would force us to take the square root of a negative number. For Denominator, we are not allowed to divide by zero. We will find where \(2x+3\) is zero and exclude it: \(x=-\frac{3}{2}\text{.}\) Thus the domain is all \(x\) with \(x\leq -1\) or \(x\geq 1\text{,}\) except for \(x=-\frac{3}{2}\text{.}\) In interval notation, this would be written as

1. For (a) we use the fact that \(g(x) = 2x-4\text{,}\) so we get the following:

   \begin{align*} 2x-4 \amp = \amp 8\\ \amp \updownarrow \amp\\ 2x\amp = \amp 12\\ \amp \updownarrow \amp\\ x\amp =\amp 6. \end{align*}

2. For (b), we use \(f(x) = x^2-3x\) and \(g(x) = 2x+4\text{.}\) Then we factor to solve:

   \begin{align*} x^2 - 3x \amp = \amp 2x-4\\ \amp \updownarrow \amp \\ x^2-5x+4 \amp = \amp 0\\ \amp \updownarrow \amp \\ (x-4)(x-1) \amp = \amp 0\\ \amp \updownarrow \amp \\ x=4 \amp \mbox{or} \amp x=1. \end{align*}

3. For (c), we need to figure out what \(f(x+1)\) means. Now \(f\) is the function that takes any input, squares it, then multiplies a new copy of it by three, then finds the difference. In this case, our input is the quantity \((x+1)\text{.}\) So we to is substitute \((x+1)\) each time we see and \(x\) in the expression for \(f(x)\) and set it equal to \(g(x)\text{:}\)

   \begin{align*} (x+1)^2 - 3(x+1) \amp = \amp 2x-4\\ \amp \updownarrow \amp \mbox{expand, group like terms on one side}\\ x^2-3x+2 \amp = \amp 0\\ \amp \updownarrow \amp \\ (x-1)(x-2) \amp = \amp 0\\ \amp \updownarrow \amp \\ x=1 \amp \mbox{or} \amp x=2. \end{align*}