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givi
20 days ago
9

If h(x) is the inverse of f(x), what is the value of h(f(x))? 0 1 x f(x) A.0 B.1 C.X D.f(x)

Mathematics
2 answers:
tester [8.8K]20 days ago
8 0

If f and h are inverse functions, then:

f (h(x) ) = x   and also

h( f(x) ) = x
tester [8.8K]20 days ago
7 0

Answer:

Correct answer is Option C

The value of h(f(x)) is x

Step-by-step explanation:

Inverse function:

An inverse function operates on another function.

For example: if f(x) is the inverse of g(x) then;

f(g(x)) = g(f(x)) = x for every x.

According to the definition:

If h(x) serves as the inverse of f(x).

Based on the theory of inverse functions:

f(h(x)) = h(f(x)) = x

Therefore, the value of h(f(x)) is x

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Question 10 Multiple Choice Worth 1 points)
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3 0
21 day ago
Which of the following is an even function g(x) = (x – 1)2 + 1g(x) = 2x2 + 1g(x) = 4x + 2g(x) = 2x
babunello [8412]
An even function can be reflected over the y-axis and still remain unchanged.
Example: y=x^2
On the other hand, an odd function can be reflected around the origin and also remains unchanged.
Example: y=x^3


A straightforward method to determine this is:

if f(x) is even, then f(-x)=f(x)
if f(x) is odd, then f(-x)=-f(x)


Hence, for an even function
substitute -x in for each and check for equivalence
make sure to fully expand the expressions
g(x)=(x-1)^2+1=x^2-2x+1+1=x^2-2x+2 is the original expression
g(x)=(x-1)^2+1
g(-x)=(-x-1)^2+1
g(-x)=(1)(x+1)^2+1
g(-x)=x^2+2x+1+1
g(-x)=x^2+2x+2
Not the same, as the original contains -2x
Therefore, it is not even
g(x)=2x^2+1
g(-x)=2(-x)^2+1
g(-x)=2x^2+1
It matches, hence it is even
g(x)=4x+2
g(-x)=4(-x)+2
g(-x)=-4x+2
Not equivalent, thus not even
g(x)=2x
g(-x)=2(-x)
g(-x)=-2x
Not equal, therefore not even



g(x)=2x²+1 is the confirmed even function.
8 0
13 days ago
Read 2 more answers
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