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QuestionFind the function (s+t)(x)(s+t)(x) for s(x)=x−2x2−25s(x)=\frac{x-2}{x^{2}-25} and t(x)=x−52−xt(x)=\frac{x-5}{2-x}. Write the domain in interval notation.

Studdy Solution

STEP 1

Assumptions1. The function s(x)s(x) is given by s(x)=x−x−25s(x)=\frac{x-}{x^{}-25}. . The function t(x)t(x) is given by t(x)=x−5−xt(x)=\frac{x-5}{-x}.
3. We are asked to find the function (s+t)(x)(s+t)(x), which is the sum of s(x)s(x) and t(x)t(x).
4. The domain of a function is the set of all possible input values (x-values) which will output real numbers.

STEP 2

We start by writing down the definition of (s+t)(x)(s+t)(x), which is the sum of s(x)s(x) and t(x)t(x).
(s+t)(x)=s(x)+t(x)(s+t)(x) = s(x) + t(x)

STEP 3

Now, we can substitute the given functions s(x)s(x) and t(x)t(x) into the equation.
(s+t)(x)=x−2x2−25+x−52−x(s+t)(x) = \frac{x-2}{x^{2}-25} + \frac{x-5}{2-x}

STEP 4

Before we can add these fractions, we need to find a common denominator. The common denominator of x2−25x^{2}-25 and 2−x2-x is (x2−25)(2−x)(x^{2}-25)(2-x).
(s+t)(x)=(x−2)(2−x)(x2−25)(2−x)+(x−)(x2−25)(x2−25)(2−x)(s+t)(x) = \frac{(x-2)(2-x)}{(x^{2}-25)(2-x)} + \frac{(x-)(x^{2}-25)}{(x^{2}-25)(2-x)}

STEP 5

Now, we can add the fractions.
(s+t)(x)=(x−2)(2−x)+(x−5)(x2−25)(x2−25)(2−x)(s+t)(x) = \frac{(x-2)(2-x) + (x-5)(x^{2}-25)}{(x^{2}-25)(2-x)}

STEP 6

Next, we simplify the numerator by distributing and combining like terms.
(s+t)(x)=2x−x2−4x+2x2+5x2−125x(x2−25)(2−x)(s+t)(x) = \frac{2x-x^{2}-4x+2x^{2}+5x^{2}-125x}{(x^{2}-25)(2-x)}

STEP 7

Combine like terms in the numerator.
(s+t)(x)=6x2−128x(x2−25)(2−x)(s+t)(x) = \frac{6x^{2}-128x}{(x^{2}-25)(2-x)}

STEP 8

The domain of the function (s+t)(x)(s+t)(x) is the set of all real numbers except the values that make the denominator equal to zero. We find these values by setting each factor of the denominator equal to zero and solving for x.
x2−25=0x^{2}-25=02−x=02-x=0

STEP 9

olve the equations for x.
For x2−25=x^{2}-25=, we get x=±5x=\pm5. For 2−x=2-x=, we get x=2x=2.

STEP 10

So, the domain of the function (s+t)(x)(s+t)(x) is all real numbers except −5-5, 22, and 55. In interval notation, this is (−∞,−5)∪(−5,2)∪(2,5)∪(5,∞)(-\infty, -5) \cup (-5,2) \cup (2,5) \cup (5, \infty).
The function (s+t)(x)(s+t)(x) is 6x2−128x(x2−25)(2−x)\frac{6x^{2}-128x}{(x^{2}-25)(2-x)} and its domain is (−∞,−5)∪(−5,2)∪(2,5)∪(5,∞)(-\infty, -5) \cup (-5,2) \cup (2,5) \cup (5, \infty).

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