Precalculus-------Give an example of three functions f,g,h none of which is a constant function such that?
fog=foh but g is not equal to h.
I tried with many functions but not getting like
f=x
g=1+x
then h=?or any other example so that we get fog=foh
To find a solution to the given problem, we need to look for functions f, g, and h that satisfy the equation fog = foh, while g ≠ h.
Let's start by defining f = x, which is a simple linear function.
Now, let's choose g = ∣x∣, which represents the absolute value of x. This function takes any input and returns the non-negative value, effectively removing the sign.
Next, we need to find h such that fog = foh. Substituting the given functions, we have ∣x∣ = h(x).
At this point, we need to find a function h(x) that doesn't have the same value as ∣x∣ for all x. A good example that fits this requirement is h(x) = x^2. When h(x) is squared, it eliminates the sign and gives a positive value for every input.
To recap:
f(x) = x
g(x) = |x|
h(x) = x^2
Now, let's check if fog equals foh:
fog = f(g(x)) = f(|x|) = |x|
foh = f(h(x)) = f(x^2) = x^2
As we can see, fog = |x| and foh = x^2. These two expressions are equal for every non-negative value of x, but not for negative values. Hence, g ≠ h.
Therefore, the functions f(x) = x, g(x) = |x|, and h(x) = x^2 satisfy the given conditions.
To find three functions f, g, and h such that fog = foh but g is not equal to h, we need to ensure that the composition of f and g is equal to the composition of f and h, while g and h remain different functions.
Let's consider the following example:
f(x) = x
g(x) = 1 + x
h(x) = 2x
Now, let's find fog(x) and foh(x):
fog(x) = f(g(x)) = f(1 + x) = 1 + x
foh(x) = f(h(x)) = f(2x) = 2x
As you can see, fog(x) is equal to foh(x).
However, g(x) = 1 + x, and h(x) = 2x, which means g and h are not the same function.
Therefore, this example satisfies the condition fog = foh while g is not equal to h.