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Mathematics 16 Online
OpenStudy (anonymous):

given f(x)= x2+3 and g(x)=x+5/x find (gof)(4)

OpenStudy (anonymous):

(gof)(x)=g(f(x)). So instead of x's in g(x) you will have f(x). Can you continue from here?

OpenStudy (anonymous):

@Traxter i still dont understand, can you work through it with me?

OpenStudy (anonymous):

(gof)(x)=g(f(x))=g(x^2+3)=(x^2+3)+5/(x^2+3) When we evaluate at x=4, we get (gof)(4)=19+5/(19)

OpenStudy (anonymous):

(Which simplifies to 366/19)

OpenStudy (anonymous):

@Traxter so its like 19.3?

OpenStudy (anonymous):

Yes, to one decimal place.

OpenStudy (anonymous):

@Traxter it cant be, its either 14/19 or 24/19

OpenStudy (anonymous):

When you wrote f(x)=x2+3, did you mean x squared by x2?

OpenStudy (anonymous):

@Traxter Yes

OpenStudy (anonymous):

@Traxter Sorry, my bad

OpenStudy (anonymous):

given f(x)= x^2+3 and g(x)=(x+5)/x find (gof)(4)

OpenStudy (anonymous):

Ok so we are taking g(x) composed with f(x) for x=4. So we will be taking g(f(4)). f(4)=4^2+3=16+3=19. Sub this into g(x). g(19)=(19+5)/19=24/19. It was the lack of brackets in g(x) that confused me. Note that x+5/x is not the same as (x+5)/x

OpenStudy (anonymous):

\[ f(x)= x^2+3\] \[g(x)=\frac{x+5}{x}\] like that?

OpenStudy (anonymous):

o then keep it x+5

OpenStudy (anonymous):

i just wanted to not that x+5 is all over x

OpenStudy (anonymous):

Then the last answer I gave is correct. Do you follow the working?

OpenStudy (anonymous):

i'm so confused then because the answers dont match up. they have to be either 14/19 24/19 15 or -15

OpenStudy (anonymous):

@Traxter

OpenStudy (anonymous):

is \(g(x)=\frac{x+5}{x}\) right?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

Yes and I told you the answer is 24/19. Look through my solution: "Ok so we are taking g(x) composed with f(x) for x=4. So we will be taking g(f(4)). f(4)=4^2+3=16+3=19. Sub this into g(x). g(19)=(19+5)/19=24/19. It was the lack of brackets in g(x) that confused me. Note that x+5/x is not the same as (x+5)/x"

OpenStudy (anonymous):

@Traxter ah i c

OpenStudy (anonymous):

@Traxter has it for sure

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