Ask your own question, for FREE!
Mathematics 20 Online
zarkam21:

Helppp

zarkam21:

1 attachment
Vocaloid:

so the function notation f(a) means "the value of the function f when a is substituted into the equation for x"

Vocaloid:

so f(2) means "the value of the function f when 2 is substituted into the equation for x"

Vocaloid:

so to find f(2) replace "x" with 2 and find f(x) = 3x^2 + 2x + 4

zarkam21:

20

Vocaloid:

good! now we replace "x" with "a+h" and find f(x) = 3x^2 + 2x + 4 your answer will not be a number, it will be a polynomial with a and h

zarkam21:

wait I got.. \[3a^2+3h^2+2a+2h+4\]

Vocaloid:

be careful (a+h)^2 is not a^2 + h^2 (this is a really common mistake though)

Vocaloid:

(a+h)^2 = (a+h)(a+h) = ? use foil to expand this

zarkam21:

what am I solving for again a or h?

Vocaloid:

you are not solving for a or h

Vocaloid:

f(a+h) just means, find f(x) when x = a + h therefore we are just plugging in "a+h" into f(x) and simplifying

Vocaloid:

3x^2 + 2x + 4 = 3(a+h)^2 + 2(a+h) + 4 = ?

zarkam21:

\[3a^2+6ah+3h^2+2a+2h+4\]

Vocaloid:

awesome, that's the answer

zarkam21:

Great !!! Now for #12

Vocaloid:

f○g = f(g(x)) to find f○g(2) we simply find g(2), take the result of that, and find f(x) of that value

Vocaloid:

so g(2) = ?

zarkam21:

17

Vocaloid:

good and f(17) = ?

zarkam21:

sorry I was afk

zarkam21:

\[\frac{ 1 }{ 17 }\]

Vocaloid:

good so that's the ans to f○g(2)

Vocaloid:

to find (f+g)(2) just add f(x) and g(x) together then let x = 2

zarkam21:

17.5

Vocaloid:

good that's it for 12

Vocaloid:

for 13 we use a similar process to 12 but we use the table instead of an equation

Vocaloid:

g(1) = ?

zarkam21:

6

Vocaloid:

good and f(6) = ?

zarkam21:

17

Vocaloid:

good that's it for 13

zarkam21:

so just 6 and 17

Vocaloid:

the answer is 17 not 6

zarkam21:

oh because g(1) is 6

Vocaloid:

yes

zarkam21:

so the final answer is 17

Vocaloid:

yes

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!