Ask your own question, for FREE!
Mathematics 16 Online
mikewwe13:

Let ​ f(x) = x + 2 ​ and ​ g(x) = 2^x + 1 ​ . Graph the functions on the same coordinate plane. What are the solutions to the equation f(x) = g(x) ? Enter your answers in the boxes. x = ____________ or x = _____________

Vocaloid:

have you tried graphing these yet?

mikewwe13:

1 attachment
Vocaloid:

good, that's correct, so what are the x-values of the solutions?

mikewwe13:

0 and 1

Vocaloid:

well done

Vocaloid:

those are your answers

mikewwe13:

wait let me get it clearer

mikewwe13:

1 attachment
mikewwe13:

What is the solution to f(x) = g(x) ? Select each correct answer. x = −2 x = −1 x = 0 x = 1 x = 2

mikewwe13:

0 and -1 are the answers correct ?

Vocaloid:

simply find each row where f(x) and g(x) have the exact same value then find the x value that matches with those rows (in the left-most column)

Vocaloid:

yes, 0 and -1 are right

mikewwe13:

1 attachment
mikewwe13:

What is the solution to f(x) = g(x) ? Select each correct answer. x= −1 x = 0 x = 1 x = 2 x = 3 x = 4 x = 5

mikewwe13:

0 and 3 are the answers ?

Vocaloid:

we are looking for the ~x~ values where f(x) and g(x) are the same

Vocaloid:

at x = 0 and x = 2 f(x) = g(x) so 0 and 2 are your answers.

mikewwe13:

https://static.k12.com/nextgen_media/assets/8090431-NG_AL1_L_01_U09_Quiz_21.png The functions f(x) = −3/4x + 2 and g(x) = (1/4)^x + 1 are shown in the graph. What are the solutions to −3/4x + 2 = (1/4)^x + 1? Select each correct answer. −1 0 1 2 3

Vocaloid:

simply find the two points where the two graphs cross each other, and identify the x-values

mikewwe13:

1 and -1

mikewwe13:

is that correct ?

Vocaloid:

try one more time

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!