Question

Consider these functions: f(x)= -2x - 1 g(x)= -1/2x + 1/2 Which statements, if any, are true about these functions? I. The function f(g(x)) = x for all real x. II. The function g(f(x)) = x for all real x. III. Functions f and g are inverse functions. A. I only B. II only C. I, II, and III D. None of the statements are true.

Answer 1

Answer:

A. I only

Step-by-step explanation:

Given the functions

h(x)= -2x - 1 g(x)= -1/2x + 1/2

h(g(x) =h( -1/2x + 1/2)

=f((-x+1)/2)

We are to replace x in h(x) with -x+1/2

h((-x+1)/2) = -2(-x+1)/2 + 1

h((-x+1)/2) = -(-x+1)+1

h((-x+1)/2)= x-1+1

h((-x+1)/2) = x

h(g(x) = x

Also for g(h(x))

g(h(x)) = g(-2x-1)

Since g(x) = -x+1/2

g(-2x-1) = -(-2x-1)+1/2

g(-2x-1) =2x+1+1/2

g(-2x-1) = 2x+2/2

g(-2x-1) = 2(x+1)/2

g(-2x-1) = x+1

Sine g(h(x)) ≠ h(g(x)), hence both functions are not inverses of each other

Hence I. The function h(g(x)) = x is the only correct option. Option A is correct

Answer 2

The correct option is: A. I only

We have functions:

$h(x)= -2x - 1 \\g(x)= -\frac{1}{2} x +\frac{1}{2} \\h(g(x)) =h(-\frac{1}{2} x +\frac{1}{2}) \\h(g(x))=f(\frac{-x+1}{2} )$

Now, replace x in h(x) by $\frac{-x+1}{2}$

So,

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

Now, g(h(x))

g(h(x)) = g(-2x-1)

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

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

As g(h(x)) ≠ h(g(x)),

Therefore, both functions are not inverses of each other.

Hence A. I only

The function h(g(x)) = x is the only correct option.

So, option A is correct.

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