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CAD60

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Can someone please help solve this problem

f(x) = 5x – 7

g(x) = 5x/(x + 4)

6(a) Write down the value of x that must be excluded from any domain of g.

6(b) Find gf(2.6)

6(c) Solve fg(x) = 2

6(d) Express the inverse function g-1(x)
 

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Part a: "... must be excluded from any domain..." is the same as saying "what values can you NOT plug into that function?"
 
Hi CAD60. I'm not familiar with the notation in parts (b) and (c).

I suppose that gf(2.6) could refer to either a composition of functions or a product of functions. Can you clarify? :)

For a composition, I've seen [imath]\color{blue}(g \circ f)(2.6)\color{black}[/imath] and
[imath]\color{blue}g(f(2.6))\color{black}[/imath].

See:
For a product, I've seen [imath]\color{blue}(g \times f)(2.6)\color{black}[/imath] and
[imath]\color{blue}(g*f)(2.6)\color{black}[/imath].

See:
 
You can't compute square roots of negative numbers and you can't divide by 0.

g(x) has no square roots----so you have no chance of computing the square root of a negative number. The denominator, x+4, equals 0 when x=? Remove that number from the domain.
 
CAD60 said:
f(x) = 5x – 7
g(x) = 5x/(x + 4)

6(a) Write down the value of x that must be excluded from any domain of g
Click to expand...
CAD60 has not responded. Here's my work for other readers.

Function g is a ratio of polynomials. All polynomials have the same domain: The set of Real numbers.

As Steven posted, the denominator in a ratio cannot be zero. Hence, the answer is -4.

CAD60 said:
6(b) Find gf(2.6)
Click to expand...
If the notation denotes function composition, then the output of function f (at x=2.6) is used as the input to function g.

f(2.6) = 5(2.6) – 7 = 6

g(6) = 5(6)/(6 + 4) = 3

Therefore: g(f(2.6)) = 3


If the notation denotes a product of functions, then the outputs of functions f and g (at x=2.6) are multiplied together.

g(2.6) = 5(2.6)/(2.6 + 4) = 1.969696…

f(2.6) [imath]\times[/imath] g(2.6) = (6)(1.969696…) = 11.818181…

CAD60 said:
6(c) Solve fg(x) = 2
Click to expand...
If the notation denotes function composition, then the output of function g is used as the input to function f.

f(g(x)) = [imath]5\big(\color{brown}\frac{5x}{x+4}\color{black}\big)-7 = \frac{25x}{x+4}-7[/imath]

Hence, the equation to solve is:

\(\displaystyle \frac{25x}{x+4}-7\,=\,2\)

Add 7 and multiply by (x+4):

25x = 9(x + 4)
25x = 9x + 36
16x = 36
x = 9/4


If the notation denotes a product of functions, then f(x) is multiplied by g(x).

\(\displaystyle (5x-7)\bigg(\frac{5x}{x+4}\bigg) = \frac{25x^{2}-35x}{x+4}\)

Hence, the equation to solve is:

\(\displaystyle \frac{25x^{2}-35x}{x+4}=2\)

Multiply by (x+4), distribute and move all terms to the left-hand side side:

\(\displaystyle 25x^{2}-37x-8 = 0\)

The quadratic formula yields two solutions:

\(\displaystyle x\,=\,\frac{37}{50}\pm\frac{3\sqrt{241}}{50}\)

CAD60 said:
6(d) Express the inverse function g-1(x)
Click to expand...
To find function g's inverse, start with y=5x/(x+4) and swap symbols x and y:

x = 5y/(y + 4)

Next, solve for y:

Multiply each side by (y+4) and distribute
Subtract 5y and 4x from each side
Factor out y on the left-hand side
Divide each side by (x–5)

g-1(x) = -4x/(x – 5)

The graphs of a function and its inverse are symmetric about the line y=x, so plotting all three with equal scales on the x- and y-axis is a good check. Here is such a graph of g(x), g-1(x) and x near the origin.

junkFMH.png
 
Otis said:
CAD60 has not responded. Here's my work for other readers.

Function g is a ratio of polynomials. All polynomials have the same domain: The set of Real numbers.

As Steven posted, the denominator in a ratio cannot be zero. Hence, the answer is -4.


If the notation denotes function composition, then the output of function f (at x=2.6) is used as the input to function g.

f(2.6) = 5(2.6) – 7 = 6

g(6) = 5(6)/(6 + 4) = 3

Therefore: g(f(2.6)) = 3


If the notation denotes a product of functions, then the outputs of functions f and g (at x=2.6) are multiplied together.

g(2.6) = 5(2.6)/(2.6 + 4) = 1.969696…

f(2.6) [imath]\times[/imath] g(2.6) = (6)(1.969696…) = 11.818181…


If the notation denotes function composition, then the output of function g is used as the input to function f.

f(g(x)) = [imath]5\big(\color{brown}\frac{5x}{x+4}\color{black}\big)-7 = \frac{25x}{x+4}-7[/imath]

Hence, the equation to solve is:

\(\displaystyle \frac{25x}{x+4}-7\,=\,2\)

Add 7 and multiply by (x+4):

25x = 9(x + 4)
25x = 9x + 36
16x = 36
x = 9/4


If the notation denotes a product of functions, then f(x) is multiplied by g(x).

\(\displaystyle (5x-7)\bigg(\frac{5x}{x+4}\bigg) = \frac{25x^{2}-35x}{x+4}\)

Hence, the equation to solve is:

\(\displaystyle \frac{25x^{2}-35x}{x+4}=2\)

Multiply by (x+4), distribute and move all terms to the left-hand side side:

\(\displaystyle 25x^{2}-37x-8 = 0\)

The quadratic formula yields two solutions:

\(\displaystyle x\,=\,\frac{37}{50}\pm\frac{3\sqrt{241}}{50}\)


To find function g's inverse, start with y=5x/(x+4) and swap symbols x and y:

x = 5y/(y + 4)

Next, solve for y:

Multiply each side by (y+4) and distribute
Subtract 5y and 4x from each side
Factor out y on the left-hand side
Divide each side by (x–5)

g-1(x) = -4x/(x – 5)

The graphs of a function and its inverse are symmetric about the line y=x, so plotting all three with equal scales on the x- and y-axis is a good check. Here is such a graph of g(x), g-1(x) and x near the origin.

View attachment 38249
Click to expand...

In my opinion another good check would be to show that
Click to expand...
g(g-1(x)) = x ...... and
g-1(g(x)) = x
 
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