Given:
f(x)=2x^2-x+1
g(x)=2 sin x
h(x)=3^x
Determine:
a) f(g(x))
b) (h^-1 o f)(x)
c) g(f(h(x))))
Please show steps, thank you very much.
Sure! Let's go step by step to determine each of the given expressions.
a) f(g(x)):
To find f(g(x)), we first need to substitute g(x) into the function f(x). Remember that g(x) represents 2sin(x).
So, substituting g(x) into f(x):
f(g(x)) = 2(g(x))^2 - g(x) + 1
Now, since g(x) = 2sin(x), we can replace g(x) in the equation:
f(g(x)) = 2(2sin(x))^2 - 2sin(x) + 1
Simplifying further:
f(g(x)) = 8sin(x)^2 - 2sin(x) + 1
Therefore, f(g(x)) = 8sin(x)^2 - 2sin(x) + 1.
b) (h^(-1) o f)(x):
To find (h^(-1) o f)(x), we need to first find the inverse of h(x) and then substitute f(x) into the inverse of h(x).
So, let's find the inverse of h(x):
To find the inverse of h(x) = 3^x, we swap x and y and solve for y.
y = 3^x
Now, swapping x and y:
x = 3^y
To solve for y, we take the logarithm base 3 of both sides:
log3(x) = log3(3^y)
log3(x) = y
Therefore, the inverse of h(x) = 3^x is h^(-1)(x) = log3(x).
Now, let's substitute f(x) = 2x^2 - x + 1 into h^(-1)(x):
(h^(-1) o f)(x) = log3(f(x))
(h^(-1) o f)(x) = log3(2x^2 - x + 1)
So, (h^(-1) o f)(x) = log3(2x^2 - x + 1).
c) g(f(h(x))):
To find g(f(h(x))), we need to substitute h(x) into f(x), and then substitute the result into g(x).
So, substituting h(x) = 3^x into f(x):
f(h(x)) = 2(3^x)^2 - 3^x + 1
Simplifying:
f(h(x)) = 2(9^x) - 3^x + 1
Now, let's substitute the result into g(x) = 2sin(x):
g(f(h(x))) = 2sin(f(h(x)))
g(f(h(x))) = 2sin(2(9^x) - 3^x + 1)
Therefore, g(f(h(x))) = 2sin(2(9^x) - 3^x + 1).
I hope this helps! Let me know if you have any further questions.