Problem

Find $ f \circ g \circ h $. $ f(x) = \tan x $ …

00:56

Problem 41 Hard Difficulty

Find $ f \circ g \circ h $.

$ f(x) = \sqrt{x - 3} $ , $ g(x) = x^2 $ , $ h(x) = x^3 + 2 $

Answer

$$\begin{aligned}
(f \circ g \circ h)(x) &=f(g(h(x)))=f\left(g\left(x^{3}+2\right)\right)=f\left[\left(x^{3}+2\right)^{2}\right] \\
&=f\left(x^{6}+4 x^{3}+4\right)=\sqrt{\left(x^{6}+4 x^{3}+4\right)-3}=\sqrt{x^{6}+4 x^{3}+1}
\end{aligned}$$

Johns Hopkins University

Watch More Solved Questions in Chapter 1

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Video Transcript

here we have functions F, g and H, and we're going to find the composition f of G of h. Another way to write that is with parentheses. F of G of age like this. So we can see from that that h of X is our innermost function and it's going to be substituted into G. And that means we're going to take a X cubed plus two and substituted in for X in the G function. And that gives us G of h of X equals X cubed plus two quantity squared. So that is G of h of X, and that whole thing gets substituted into the F function. So that whole thing goes in for X in the F function. And that's going to result in the square root of X cubed plus two quantity squared minus three. Now perhaps we want to simplify that. And if we do, we're going to end up squaring this X cubed plus two. We're going to multiply it out using the foil method, and that would give us X to the six power plus four X to the third power plus four and then we still have the minus three. So the last thing we want to do is combine the like terms and we have at the square root of X to the sixth power plus four x to the third power plus one.

Heather Z.

Oregon State University

Related Courses

Calculus 1 / AB

Calculus 2 / BC

Calculus 3

Calculus: Early Transcendentals

Chapter 1

Functions and Models

Section 3

New Functions from Old Functions

Problem

Find $ f \circ g \circ h $. $ f(x) = \tan x $ …

Problem 41 Hard Difficulty

Find $ f \circ g \circ h $.

$ f(x) = \sqrt{x - 3} $ , $ g(x) = x^2 $ , $ h(x) = x^3 + 2 $

Answer

$$\begin{aligned}
(f \circ g \circ h)(x) &=f(g(h(x)))=f\left(g\left(x^{3}+2\right)\right)=f\left[\left(x^{3}+2\right)^{2}\right] \\
&=f\left(x^{6}+4 x^{3}+4\right)=\sqrt{\left(x^{6}+4 x^{3}+4\right)-3}=\sqrt{x^{6}+4 x^{3}+1}
\end{aligned}$$

More Answers

Related Courses

Calculus: Early Transcendentals

Related Topics

Discussion

Top Calculus 3 Educators

Lily A.

Caleb E.

Samuel H.

Joseph L.

Calculus 3 Courses

Vectors Intro

Vector Basics - Example 1

Recommended Videos

Watch More Solved Questions in Chapter 1

Video Transcript

Calculus: Early Transcendentals

Lily A.

Caleb E.

Samuel H.

Joseph L.

Vectors Intro

Vector Basics - Example 1

$$\begin{aligned}(f \circ g \circ h)(x) &=f(g(h(x)))=f\left(g\left(x^{3}+2\right)\right)=f\left[\left(x^{3}+2\right)^{2}\right] \\&=f\left(x^{6}+4 x^{3}+4\right)=\sqrt{\left(x^{6}+4 x^{3}+4\right)-3}=\sqrt{x^{6}+4 x^{3}+1}\end{aligned}$$

Calculus: Early Transcendentals

Lily A.

Caleb E.

Samuel H.

Joseph L.

Vectors Intro

Vector Basics - Example 1

Lily A.

Caleb E.

Samuel H.

Joseph L.

$$\begin{aligned}
(f \circ g \circ h)(x) &=f(g(h(x)))=f\left(g\left(x^{3}+2\right)\right)=f\left[\left(x^{3}+2\right)^{2}\right] \\
&=f\left(x^{6}+4 x^{3}+4\right)=\sqrt{\left(x^{6}+4 x^{3}+4\right)-3}=\sqrt{x^{6}+4 x^{3}+1}
\end{aligned}$$