Question

How would I do this??

Answer 1

Part 1

$\left(\frac{g}{h}\right)(x) = \frac{g(x)}{h(x)}\\\\\left(\frac{g}{h}\right)(x) = \frac{3x-5}{-2x^2+7}\\\\\left(\frac{g}{h}\right)(3) = \frac{3(3)-5}{-2(3)^2+7}\\\\\left(\frac{g}{h}\right)(3) = \frac{4}{-11}\\\\\left(\frac{g}{h}\right)(3) = -\frac{4}{11}$

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.

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

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: $x = \sqrt{3.5} \text{ and } x = -\sqrt{3.5}$