Precalculus-------Give an example of three functions f,g,h none of which is a constant function such that?
Precalculus-------Give an example of three functions f,g,h none of which is a constant function such that?
foh=goh but f is not equal to g
I tried with many functions but not getting like
f=x
g=1+x^2
then h=?or any other example so that we get foh=goh
To find an example of three functions f, g, and h such that f∘h = g∘h but f ≠ g, we can start with two functions f and g and then find a suitable h function.
Let's take f(x) = x and g(x) = 1 + x^2, as you already mentioned. To find h(x), we need to find a function that maps the domain of g (which is all real numbers) to the domain of f (also all real numbers) in such a way that f∘h = g∘h.
Notice that g(x) is a quadratic function that maps every real number x to a unique value. So we need h(x) to map each value of x from the domain of g to a unique value in the domain of f.
One approach could be to use a linear function for h(x). Let's try h(x) = 2x.
Now, we need to verify that f∘h = g∘h. We can do this by evaluating both sides of the equation.
(f∘h)(x) = f(h(x)) = f(2x) = 2x
(g∘h)(x) = g(h(x)) = g(2x) = 1 + (2x)^2 = 1 + 4x^2
As you can see, (f∘h)(x) = 2x is not equal to (g∘h)(x) = 1 + 4x^2. Therefore, h(x) = 2x is not a suitable function that satisfies f∘h = g∘h for the given f(x) and g(x).
Let's try another approach. Let's take h(x) = √(x - 1).
Now, we need to verify that f∘h = g∘h. Again, we evaluate both sides of the equation.
(f∘h)(x) = f(h(x)) = f(√(x - 1)) = √(x - 1)
(g∘h)(x) = g(h(x)) = g(√(x - 1)) = 1 + (√(x - 1))^2 = 1 + (x - 1) = x
Now, we have (f∘h)(x) = √(x - 1) and (g∘h)(x) = x. By comparing the two, we can see that f∘h = g∘h, but f ≠ g.
Therefore, one example of functions satisfying f∘h = g∘h but f ≠ g is:
f(x) = x
g(x) = 1 + x^2
h(x) = √(x - 1)