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Graphs and Functions
Combining functions: Advanced
Suppose that the functions
f and
g are defined as follows.

{:[f(x)=sqrt(4x-5)],[g(x)=x-4]:}
Find
f*g and
f+g. Then, give their domains using interval notation.

(f*g)(x)=

◻
Domain of
f*g:
◻

(f+g)(x)=

◻
Domain of
f+g:
◻

Graphs and Functions $\newline$ Combining functions: Advanced $\newline$ Suppose that the functions $f$ and $g$ are defined as follows. $\newline$ $\begin{array}{l} f(x)=\sqrt{4 x-5} \\ g(x)=x-4 \end{array}$ $\newline$ Find $f \cdot g$ and $f+g$ . Then, give their domains using interval notation. $\newline$ $(f \cdot g)(x)=$ $\newline$ $\square$ $\newline$ Domain of $f \cdot g:$ $\square$ $\newline$ $(f+g)(x)=$ $\newline$ $\square$ $\newline$ Domain of $f+g:$ $\square$

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Compare linear and exponential growth

Full solution

Q. Graphs and Functions

\newline

Combining functions: Advanced

\newline

Suppose that the functions

f

and

g

are defined as follows.

\newline

\begin{array}{l} f(x)=\sqrt{4 x-5} \\ g(x)=x-4 \end{array}

\newline

Find

f \cdot g

and

f+g

. Then, give their domains using interval notation.

\newline

(f \cdot g)(x)=

\newline

\square

\newline

Domain of

f \cdot g:

\square

\newline

(f+g)(x)=

\newline

\square

\newline

Domain of

f+g:

\square

Define functions and operations: Step $1$ : Define the functions and identify the operations needed. $\newline$ $f(x) = \sqrt{4x - 5}$ $\newline$ $g(x) = x - 4$ $\newline$ We need to find $(f*g)(x)$ and $(f+g)(x)$ .
Calculate $(f*g)(x)$ : Step $2$ : Calculate $(f*g)(x) = f(x) * g(x)$ . $(f*g)(x) = \sqrt{4x - 5} * (x - 4)$ Simplify if possible, but here it remains as is.
Determine domain of $(f*g)(x)$ : Step $3$ : Determine the domain of $(f*g)(x)$ . $\newline$ The domain of $\sqrt{4x - 5}$ is $4x - 5 \geq 0$ , so $x \geq \frac{5}{4}$ . $\newline$ The domain of $x - 4$ is all real numbers. $\newline$ The domain of $(f*g)(x)$ is the intersection, which is $x \geq \frac{5}{4}$ .
Calculate $(f+g)(x)$ : Step $4$ : Calculate $(f+g)(x) = f(x) + g(x)$ . $(f+g)(x) = \sqrt{4x - 5} + (x - 4)$ Again, this expression is simplified as much as possible.
Determine domain of $(f+g)(x)$ : Step $5$ : Determine the domain of $(f+g)(x)$ . The domain of $\sqrt{4x - 5}$ is $4x - 5 \geq 0$ , so $x \geq \frac{5}{4}$ . The domain of $x - 4$ is all real numbers. The domain of $(f+g)(x)$ is the intersection, which is $x \geq \frac{5}{4}$ .

More problems from Compare linear and exponential growth

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x

days to pick up the motorcycle.

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Write an equation for the function. If it is linear, write it in the form

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f(t)=-2t-7

change over the interval from

t=-7

t=-6

\newline

\newline

\left[\text{f(t) decreases by } 2\right]

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\left[\text{f(t) increases by } 2\right]

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g(t)=9^t

change over the interval from

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\left[\text{g(t) decreases by a factor of } 81\right]

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\left[\text{g(t) increases by a factor of } 81\right]

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Question

Both of these functions grow as

x

gets larger and larger. Which function eventually exceeds the other?

\newline

\newline

(A)f(x) = 8x + 3.3

\newline

(B)g(x) = 3.3^x - 5

Posted 6 months ago

Question

Both of these functions grow as

x

gets larger and larger. Which function eventually exceeds the other?

\newline

\newline

(A) \ f(x) = 2x + 9

\newline

(B)\ g(x) = 2^x

Posted 6 months ago

Question

f(x)

g(x)

are differentiable.

\newline

h(x)

h(x)=\frac{g(x)}{f(x)}

\newline

f(8)=1

f'(8)=-6

g(8)=8

g'(8)=-10

h'(8)

\newline

Simplify any fractions.

\newline

h'(8)=

Posted 6 months ago

Question

p(x)

q(x)

are differentiable.

\newline

r(x)

r(x)= \frac{p(x)}{q(x)}

\newline

p(3)= 2

p'(3)= 4

q(3)= 6

q'(3)= 9

r'(3)

\newline

Simplify any fractions.

\newline

r'(3)=

Posted 7 months ago

Question

u(x)

v(x)

are differentiable.

\newline

w(x)

w(x)= \frac{u(x)}{v(x)}

\newline

u(5)= 3

u'(5)= -2

v(5)= 7

v'(5)= 1

w'(5)

\newline

Simplify any fractions.

\newline

w'(5)=

Posted 7 months ago

Question

a(x)

b(x)

are differentiable.

\newline

c(x)

c(x)= \frac{a(x)}{b(x)}

\newline

a(2)= 4

a'(2)= 5

b(2)= 1

b'(2)= -3

c'(2)

\newline

Simplify any fractions.

\newline

c'(2)=

Posted 7 months ago

Question

m(x)

n(x)

are differentiable.

\newline

o(x)

o(x)= \frac{m(x)}{n(x)}

\newline

m(7)= 2

m'(7)= -1

n(7)= 4

n'(7)= 8

o'(7)

\newline

Simplify any fractions.

\newline

o'(7)=

Posted 7 months ago

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