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2 years ago

If f(x) = x2 - 1, and g(x) = x + 2, then f(g(x)) = [? ]x2+[?]x+[?]

Mathematics

1 answer:

klio [65]2 years ago

4 0

Step-by-step explanation:

f.g(x) = f(x+2)

=(x+2)2 -1

=(x2+2*x*2+4)-1

=x2+4x+4-1

=x2+4x+3

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Solve:<br><br> a.x3+x2-8x - 12x+2<br> b. x3-4x2-3x + 18x-3<br> c. x2 + 4x + 4<br> d. x2 - 6x + 9

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<h2>Steps:</h2>

So firstly, this function is asking us to divide f(x) by g(x) so let's set that up as such:

$(\frac{f}{g})(x)=\frac{x^2-x-6}{1}\div \frac{x-3}{x+2}$

Next, remember that <u>dividing by a number is the same as multiplying by its reciprocal.</u> To find the reciprocal of a number, flip the numerator and denominator around. With this info, flip the second fraction to it's reciprocal and change the sign to multiplication:

$(\frac{f}{g})(x)=\frac{x^2-x-6}{1}\times \frac{x+2}{x-3}$

Next, we are going to factor x² - x - 6. Firstly, what two terms have a product of -6x² and a sum of -x? That would be -3x and 2x. Replace -x with 2x - 3x:

$(\frac{f}{g})(x)=\frac{x^2+2x-3x-6}{1}\times \frac{x+2}{x-3}$

Next, factor x² + 2x and -3x - 6 separately. Make sure that they have the same quantity on the inside of the parentheses:

$(\frac{f}{g})(x)=\frac{x(x+2)-3(x+2)}{1}\times \frac{x+2}{x-3}$

Now you can rewrite it as:

$(\frac{f}{g})(x)=\frac{(x-3)(x+2)}{1}\times \frac{x+2}{x-3}$

Next, multiply:

$(\frac{f}{g})(x)=\frac{(x-3)(x+2)^2}{x-3}$

Next, divide:

$(\frac{f}{g})(x)=(x+2)^2$

And lastly, simplify:

A good tip: $(x+y)^2=x^2+2xy+y^2$

$(\frac{f}{g})(x)=x^2+4x+4$

<h2>Answer:</h2>

<u>The correct option is C. x² + 4x + 4.</u>

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