Question

Solve the simpler problems first, and check your work if time permits. 1a. What is the domain of the function defined by F(x)=x^2−25/x^2+3x−10​ ? Show your work below. Domain = b. Write the above fraction in simplest form.

Answer 1

a. The domain of F(x) is:

`\[ \text{Domain} = \{ x \in \mathbb{R} \mid x \\eq -5, x \\eq 2 \} \]`

b.
`\[ F(x) = (x - 5)/(x - 2) \]`

Step-by-step explanation:

Let's solve both parts of this question:

Part a: Finding the Domain of F(x)

The domain of the function
`\( F(x) = (x^2 - 25)/(x^2 + 3x - 10) \)` will include all real numbers except those that make the denominator equal to zero because division by zero is undefined.

To find which values are not in the domain, we need to solve for x in the denominator:

`\[ x^2 + 3x - 10 = 0 \]`

This is a quadratic equation that can be factored or solved using the quadratic formula. We will attempt to factor it first. The factors of -10 that add up to 3 are 5 and -2.

By factoring, we have:

`\[ (x + 5)(x - 2) = 0 \]`

Setting each factor equal to zero gives us the excluded values:

`\[ x + 5 = 0 \Rightarrow x = -5 \]`

`\[ x - 2 = 0 \Rightarrow x = 2 \]`

These are the values for which the denominator becomes zero, and therefore, they must be excluded from the domain.

Hence, the domain of F(x) is:

`\[ \text{Domain} = \{ x \in \mathbb{R} \mid x \\eq -5, x \\eq 2 \} \]`

Part b: Simplifying the Fraction

The function F(x) is defined by:

`\[ F(x) = (x^2 - 25)/(x^2 + 3x - 10) \]`

Both the numerator and the denominator are quadratic expressions. We can factor both expressions as they are differences of squares and a simple quadratic, respectively.

The numerator is a difference of squares and can be factored as:

`\[ x^2 - 25 = (x + 5)(x - 5) \]`

As we factored the denominator in part a, it is:

`\[ x^2 + 3x - 10 = (x + 5)(x - 2) \]`

Now we write the fraction in factored form:

`\[ F(x) = ((x + 5)(x - 5))/((x + 5)(x - 2)) \]`

We see that the factor (x + 5) appears in both the numerator and the denominator. Since we are not including x = -5 in our domain, we can cancel out this common factor:

`\[ F(x) = (x - 5)/(x - 2) \]`

This is the simplest form of the fraction F(x) with the restriction that
`\( x \\eq -5 \) and \( x \\eq 2 \)` due to the domain consideration.