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3 years ago

Determining whether two functions are inverses of each other.

Mathematics

2 answers:

Gnom [1K]3 years ago

8 0

Answer:

f and g are not inverse of each other.

Step-by-step explanation:

If the two functions are inverses of each other, then f(g(x)) = g(f(x))

We are given two functions:

f(x) = 6x + 5

g(x) = 6x - 5

Let's find f(g(x)) and g(f(x))

f(g(x)) = f(6x - 5)

= 6(6x - 5) + 5

= 36x - 30 + 5

f(g(x)) = 36x - 25

Now let's find g(f(x))

g(f(x)) = g(6x + 5)

= 6(6x + 5) - 5

= 36x + 30 - 5

g(f(x))= 36x + 25

We see that f(g(x)) ≠ g(f(x))

Therefore, f and g are not inverse of each other.

murzikaleks [220]3 years ago

4 0

F(x) = 6x + 5
g(x) = 6x - 5
f(g(x)) = 6(6x - 5) + 5 = 36x - 30 + 5 = 36x - 25
g(f(x)) = 6(6x+ 5) - 5 = 36x + 30 - 5 = 36x + 25

f and g are not inverse of each other.

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Explain how to solve 5^(x − 2) = 8 using the change of base formula log base b of y equals log y over log b.

Ray Of Light [21]

Answer:

x = 3.3

Step-by-step explanation:

A equation is given to us and we need to solve out for x. The given equation is ,

Take log on both sides with base as " 10" . We have ,

Simplify using the property of log , , we have ,

Simplify ,

Again simplify using the property of log ,

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2 years ago

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3 years ago

Which of the following values would complete the ordered pair if the point is on the graph of f (x) = -3x + 2 ?

Scrat [10]

Answer: the y-coordinate that completes the point is 5.

Explanation:

An ordered pair (x, y) indicates the x-and-y coordinates of a general point, where x is the input value of a function and y is the output value.

The output value is y = f(x) and must be found applying the rule (function) to the given input value.

In this case the rulte is f(x) = - 3x + 2, and the input value, x, is - 1 (the first value of the ordered pair).

This is the mathematical procedure:

x = - 1
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2 years ago

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Free_Kalibri [48]

QUESTION 1

We want to solve,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}$

We factor the denominator of the fraction on the right hand side to get,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.$

This implies
$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.$

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}$

We multiply through by LCM of
$(x-4)(x - 2)$

$(x - 2) + x(x-4) = 2$

We expand to get,

$x - 2 + {x}^{2} - 4x= 2$

We group like terms and equate everything to zero,

${x}^{2} + x - 4x - 2 - 2 = 0$

We split the middle term,

${x}^{2} + - 3x - 4 = 0$

We factor to get,

${x}^{2} + x - 4x- 4 = 0$

$x(x + 1) - 4(x + 1) = 0$

$(x + 1)(x - 4) = 0$

$x + 1 = 0 \: or \: x - 4 = 0$

$x = - 1 \: or \: x = 4$

But
$x = 4$
is not in the domain of the given equation.

It is an extraneous solution.

$\therefore \: x = - 1$
is the only solution.

QUESTION 2

$\sqrt{x+11} -x=-1$

We add x to both sides,

$\sqrt{x+11} =x-1$

We square both sides,

$x + 11 = (x - 1)^{2}$

We expand to get,

$x + 11 = {x}^{2} - 2x + 1$

This implies,

${x}^{2} - 3x - 10 = 0$

We solve this quadratic equation by factorization,

${x}^{2} - 5x + 2x - 10 = 0$

$x(x - 5) + 2(x - 5) = 0$

$(x + 2)(x - 5) = 0$

$x + 2 = 0 \: or \: x - 5 = 0$

$x = - 2 \: or \: x = 5$

But
$x = - 2$
is an extraneous solution

$\therefore \: x = 5$

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2 years ago

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