Question

calculus

Answer 1

@dude

Answer 2

@vocaloid

Answer 3

@smokeybrown

Answer 4

@thesmartone

Answer 5

It seems like an interesting question, but I am having some trouble understanding the formatting.
The equations for f(x) and g(x) are given as 2x and x+1, respectively; the question then asks that we choose the graph matching
\[(F+G)^{-1}(x)\]
of those two equations.

For starters, I think it would be good to show the other choices of graphs that we can select. We also need to clarify the meaning of capital F and G and their relation to lowercase f and g

Answer 6

A.

Answer 7

B^^

Answer 8

C

Answer 9

D

Answer 10

E

Answer 11

F

Answer 12

G

Answer 13

**SORRY FOR NOT POSTING GRAPHS, I FOGOT BUT THATS ALL**

Answer 14

Would it be safe to assume in this case that F(x) and G(x) represent antiderivatives of f(x) and g(x)?
If that is the case, then F(x) would resolve to x^2 and G(x) would resolve to 1/2(x^2) + x, I think.
Then, (F+G) (x) would be 3/2(x^2) + x,
And finally
(F+G)^-1(x) would be the inverse of the previous equation, which I am unfortunately not sure how to calculate, whoops

I suppose we could try to plug the inverse-equation into a graphing calculator and compare the result to our options, but maybe that would defeat the purpose of the problem?

Answer 15

\( \left(f+g\right)^{-1}\left(x\right),\:f\left(x\right)=2x,\:g\left(x\right)=x+1 \)

\( \mathrm{first let's find\:}\left(f+g\right)^{-1}\left(x\right)\:\mathrm{given}\:f\left(x\right)=2x,\:g\left(x\right)=x+1 \)

that would give us \( \left(2x+x+1\right)^{-1} \)


so the function we have so far is \( \left(2x+x+1\right)^{-1}x \) which is the inverse basically -- we can apply the rule of negatative power of one:

\( \frac{1}{3x+1}x \ \ \vee \ \ \frac{x}{3x+1} \)

i do not want to get into graphing formulas so just used desmos, great website, and got the image attached; now its up to you to pick the answer!!! : )

Answer 16

Thank you Guys, may i have the Desmos Link?

Answer 17

https://www.desmos.com/calculator/xeprkvut5i

Answer 18

Thank you So much