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.

a)

f(g(x))
= f(2sinx)
= 2(4sin^2 x) - 2sinx + 1
= 8sin^2 x - 2sinx + 1

for b) find the inverse of 3^x first
y = 3^x
inverse is x = 3^y
lnx = y ln3
y = (lnx)/ln3
so (h^-1 o f)(x) = ln(2x^2 - x + 1)/ln3

you try the last

2sin(2(3^x)^2 - 3^x + 1)

2sin(2(3^2x) - 3^x + 1) ??

correct!

good job

To solve these problems, we first need to understand the concept of function composition. Function composition is when we combine two or more functions to create a new function.

a) To find f(g(x)), we substitute the function g(x) into the function f(x).

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

Substituting g(x) = 2sin(x):

f(g(x)) = 2(2sin(x))^2 - 2sin(x) + 1

Simplifying:

f(g(x)) = 8sin^2(x) - 2sin(x) + 1

So, f(g(x)) = 8sin^2(x) - 2sin(x) + 1.

b) To find (h^ -1 o f)(x), we need to find the inverse of h(x) first.

To find the inverse of h(x) = 3^x, we switch the x and y and solve for y.

Let y = 3^x:

x = 3^y

To solve for y, take the logarithm base 3 of both sides:

log3(x) = log3(3^y)

log3(x) = y

Therefore, the inverse of h(x) is h^-1(x) = log3(x).

Now we substitute f(x) into h^-1(x):

(h^-1 o f)(x) = log3(f(x))

Substituting f(x) = 2x^2 - x + 1:

(h^-1 o f)(x) = log3(2x^2 - x + 1)

So, (h^-1 o f)(x) = log3(2x^2 - x + 1).

c) To find g(f(h(x))), we substitute h(x) into f(x) and g(x) into f(x).

g(f(h(x))) = g(f(3^x))

Substituting f(x) = 2x^2 - x + 1:

g(f(h(x))) = g(2(3^x)^2 - 3^x + 1)

Simplifying:

g(f(h(x))) = g(2(9^x) - 3^x + 1)

Finally, substituting g(x) = 2sin(x):

g(f(h(x))) = 2sin(2(9^x) - 3^x + 1)

Therefore, g(f(h(x))) = 2sin(2(9^x) - 3^x + 1).