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Numerade Educator



Problem 31 Hard Difficulty

Find (a) $ f + g $, (b) $ f - g $, (c) $ fg $ and (d) $ f/g $ and state their domains.

$ f(x) = x^3 + 2x^2 $ , $ g(x) = 3x^2 - 1 $


$f(x)=x^{3}+2 x^{2} ; g(x)=3 x^{2}-1, \quad D=\mathbb{R}$ for both $f$ and $g$.
(a) $(f+g)(x)=\left(x^{3}+2 x^{2}\right)+\left(3 x^{2}-1\right)=x^{3}+5 x^{2}-1, D=(-\infty, \infty),$ or $\mathbb{R}$.
(b) $(f-g)(x)=\left(x^{3}+2 x^{2}\right)-\left(3 x^{2}-1\right)=x^{3}-x^{2}+1, D=\mathbb{R}$.
(c) $(f g)(x)=\left(x^{3}+2 x^{2}\right)\left(3 x^{2}-1\right)=3 x^{3}+6 x^{4}-x^{3}-2 x^{2}, D=\mathbb{R}$.
(d) $\left(\frac{f}{g}\right)(x)=\frac{x^{3}+2 x^{2}}{3 x^{2}-1}, D=\left\{x | x \neq \pm \frac{1}{\sqrt{3}}\right\}$ since $3 x^{2}-1 \neq 0$.

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Video Transcript

here we have functions. F N g. So let's start by finding F plus g. So we're going to add these functions together. X cubed plus two X squared plus three X squared minus one. We can add the like terms and we get X cubed plus five X squared minus one Breath plus G for F minus. G will subtract them. So we have X cubed plus two X squared, minus the quantity three X squared minus one. And so let's distribute the negative sign to have X cubed plus two X squared minus three X squared, plus one. And then let's combine like terms and we have X cubed minus X squared plus one and thats f minus g. Now let's look at the domains so F and G or both polynomial functions and polynomial functions always have a domain of all real numbers, which we can write as negative infinity to infinity. And when you add or subtract functions, the domain of the new function will be just the intersection of the two separate functions, and the intersection of all real numbers and all real numbers is all real numbers. All right, now, let's take a look at the product and the quotient. So if we multiply f and G X cubed plus two x squared multiplied by three X squared minus one, we might want to multiply that out so we'll use the foil process will multiply the first and we get three X to the fifth Power the outsides and we get minus X cubed the insides and we get plus six x to the fourth power and the last time I get minus two x squared. And if we want to rewrite that in descending powers of X, we get half times X equals three x to the fifth power plus six x to the fourth power minus X cubed minus two X squared. Now let's find the quotient of F over G. So we have X cubed plus two x squared over three X squared minus one. That's about all we can do with that one. Now let's talk about the domains. So again, the domain of F was all real numbers, and the domain of G was all real numbers. And when you multiply them, you still get the intersection of the domains. So you get all real numbers again. Now something changes when we divide, though, because we can't divide by zero. We need three X squared minus one to not be equal to zero. So what would make that equal zero if three X squared equaled one. So if X squared equals 1/3 So if X was plus or minus the square root of 1/3 then we would have a problem. We would have zero on the bottom. So we're going to say, for the domain is all real numbers except X equals plus or minus square 1/3. So I'll just say X is not equal to plus or minus square root 1/3.