Composition of Functions

In ...

Question

Composition of Functions

In Exercises 39 and 40, find

a. (ƒ ○ g) (-1).

ƒ(x) = 1/x , g(x) = 1/√ x + 2

Accepted Answer

Welcome back, everyone. Suppose G of X equals 2 divided by X and H of X equals 1 divided by square root of X minus 3. Find the composition G H 7. We're given 4 answer choices A2, B4, C 11/2, and D 14. So let's understand the given notation. It says G composition of H at 0.7. So we can basically rewrite it as G of H of 7, right? Essentially, this means that instead of our original variable X, we are replacing it with the function H. In a general case, if we want to identify the composition G of H of X, this means that we're getting G of H of X. So now for every X that we have in G of X, we're going to replace it with H. So 2 divided by X now becomes G H equals 2 divided by H, right? And now, since H is a function of X, we're going to get 2 divided by H of X. And now, if we rewrite H of X in terms of X, well, H of X is defined as 1 divided by square root of X minus 3. So G of H of X is simply equal to. 2 square root of x minus 3. And now if we want to value G of H of 7, we are simply substituting X equals 7, which gives us 2 square root of 7 minus 3. And this is equal to 2 multiplied by square root of 4 witches. 2 multiplied by 2 and that's 4. So the final answer to this problem corresponds to the answer choice B. Thank you for watching.

Composition of Functions

In Exercises 39 and 40, find

a. (ƒ ○ g) (-1).

ƒ(x) = 1/x , g(x) = 1/√ x + 2

Watch next

Composition of FunctionsIn Exercises 39 and 40, finda. (ƒ ○ g) (-1).ƒ(x) = 1/x , g(x) = 1/√ x + 2

Watch next

Composition of Functions

In Exercises 39 and 40, find

a. (ƒ ○ g) (-1).

ƒ(x) = 1/x , g(x) = 1/√ x + 2