Math

QuestionFind (s+t)(x)(s+t)(x) for s(x)=x4x225s(x)=\frac{x-4}{x^{2}-25} and t(x)=x54xt(x)=\frac{x-5}{4-x}, and state the domain.

Studdy Solution

STEP 1

Assumptions1. The functions are s(x)=x4x25s(x)=\frac{x-4}{x^{}-25} and t(x)=x54xt(x)=\frac{x-5}{4-x}. . We need to find the function (s+t)(x)(s+t)(x), which is the sum of the functions s(x)s(x) and t(x)t(x).
3. The domain of a function is the set of all possible input values (x-values) which will output real numbers.

STEP 2

To find the function (s+t)(x)(s+t)(x), we simply add the functions s(x)s(x) and t(x)t(x) together.
(s+t)(x)=s(x)+t(x)(s+t)(x) = s(x) + t(x)

STEP 3

Substitute the given functions s(x)s(x) and t(x)t(x) into the equation.
(s+t)(x)=xx225+x5x(s+t)(x) = \frac{x-}{x^{2}-25} + \frac{x-5}{-x}

STEP 4

In order to add these two fractions, we need to find a common denominator. The common denominator of x225x^{2}-25 and 4x4-x is (x225)(4x)(x^{2}-25)(4-x).
So, we rewrite the fractions with the common denominator.
(s+t)(x)=(x4)(4x)(x225)(4x)+(x)(x225)(x225)(4x)(s+t)(x) = \frac{(x-4)(4-x)}{(x^{2}-25)(4-x)} + \frac{(x-)(x^{2}-25)}{(x^{2}-25)(4-x)}

STEP 5

Combine the two fractions into one.
(s+t)(x)=(x4)(4x)+(x5)(x225)(x225)(4x)(s+t)(x) = \frac{(x-4)(4-x) + (x-5)(x^{2}-25)}{(x^{2}-25)(4-x)}

STEP 6

implify the numerator by distributing and combining like terms.
(s+t)(x)=4xx216+4xx3+25x125(x225)(4x)(s+t)(x) = \frac{4x - x^{2} -16 +4x - x^{3} +25x -125}{(x^{2}-25)(4-x)}(s+t)(x)=x3+3x2+33x141(x225)(4x)(s+t)(x) = \frac{-x^{3} +3x^{2} +33x -141}{(x^{2}-25)(4-x)}

STEP 7

The domain of the function (s+t)(x)(s+t)(x) is all real numbers except the values that make the denominator equal to zero.To find these values, we set each factor of the denominator equal to zero and solve for xx.
x225=0x^{2}-25=04x=04-x=0

STEP 8

olve the equations to find the values of xx that are not in the domain.
x225=0x=±5x^{2}-25=0 \Rightarrow x=\pm54x=0x=44-x=0 \Rightarrow x=4

STEP 9

So, the domain of the function (s+t)(x)(s+t)(x) is all real numbers except 5-5, 44, and 55.In interval notation, this is (,5)(5,4)(4,5)(5,)(-\infty, -5) \cup (-5,4) \cup (4,5) \cup (5, \infty).
The function (s+t)(x)(s+t)(x) is(s+t)(x)=x3+3x2+33x141(x225)(4x)(s+t)(x) = \frac{-x^{3} +3x^{2} +33x -141}{(x^{2}-25)(4-x)}And the domain is (,5)(5,4)(4,5)(5,)(-\infty, -5) \cup (-5,4) \cup (4,5) \cup (5, \infty).

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