Question

How would I do this??

Answer 1

Part 1

`\left((g)/(h)\right)(x) = (g(x))/(h(x))\\\\\left((g)/(h)\right)(x) = (3x-5)/(-2x^2+7)\\\\\left((g)/(h)\right)(3) = (3(3)-5)/(-2(3)^2+7)\\\\\left((g)/(h)\right)(3) = (4)/(-11)\\\\\left((g)/(h)\right)(3) = -(4)/(11)\\\\`

Answer: -4/11

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Part 2

Set the denominator function equal to zero and solve for x to find which values to kick out of the domain.

`h(x) = 0\\\\-2x^2+7 = 0\\\\7 = 2x^2\\\\2x^2 = 7\\\\x^2 = 7/2\\\\x^2 = 3.5\\\\x = √(3.5) \ \text{ or } x = -√(3.5)\\\\`

This shows that if x is equal to either of those values, then the denominator h(x) will be zero. These are the values to kick out of the domain to prevent a division by zero error. Any other value of x is valid in the domain.

Answer:
`x = √(3.5) \text{ and } x = -√(3.5)\\\\`