The given functions are $f(x) = 2x - 1$ and $g(x) = x^2 - 2x + 1$. We need to find the following:

1. $(f \circ g)(x)$
2. $(g \circ f)(x)$
3. $(f \circ f)(-1)$
4. $(g \circ g)(1)$
5. $(f \circ g)(1.5)$

Let's solve these step by step:

1. $(f \circ g)(x)$

This means $f(g(x))$.

First, calculate $g(x)$: $g(x) = x^2 - 2x + 1$

Now, substitute $g(x)$ into $f$: $f(g(x)) = f(x^2 - 2x + 1)$ $f(x^2 - 2x + 1) = 2(x^2 - 2x + 1) - 1$ $= 2x^2 - 4x + 2 - 1$ $= 2x^2 - 4x + 1$

Thus, $(f \circ g)(x) = 2x^2 - 4x + 1$

2. $(g \circ f)(x)$

This means $g(f(x))$.

First, calculate $f(x)$: $f(x) = 2x - 1$

Now, substitute $f(x)$ into $g$: $g(f(x)) = g(2x - 1)$ $g(2x - 1) = (2x - 1)^2 - 2(2x - 1) + 1$ $= (2x - 1)^2 - 4x + 2 + 1$ $= 4x^2 - 4x + 1 - 4x + 2 + 1$ $= 4x^2 - 8x + 4$

Thus, $(g \circ f)(x) = 4x^2 - 8x + 4$

3. $(f \circ f)(-1)$

This means $f(f(-1))$.

First, calculate $f(-1)$: $f(-1) = 2(-1) - 1$ $= -2 - 1$ $= -3$

Now, substitute $-3$ into $f$: $f(-3) = 2(-3) - 1$ $= -6 - 1$ $= -7$

Thus, $(f \circ f)(-1) = -7$

4. $(g \circ g)(1)$

This means $g(g(1))$.

First, calculate $g(1)$: $g(1) = 1^2 - 2(1) + 1$ $= 1 - 2 + 1$ $= 0$

Now, substitute $0$ into $g$: $g(0) = 0^2 - 2(0) + 1$ $= 0 + 0 + 1$ $= 1$

Thus, $(g \circ g)(1) = 1$

5. $(f \circ g)(1.5)$

This means $f(g(1.5))$.

First, calculate $g(1.5)$: $g(1.5) = (1.5)^2 - 2(1.5) + 1$ $= 2.25 - 3 + 1$ $= 0.25$

Now, substitute $0.25$ into $f$: $f(0.25) = 2(0.25) - 1$ $= 0.5 - 1$ $= -0.5$

Thus, $(f \circ g)(1.5) = -0.5$

Summary:

1. $(f \circ g)(x) = 2x^2 - 4x + 1$
2. $(g \circ f)(x) = 4x^2 - 8x + 4$
3. $(f \circ f)(-1) = -7$
4. $(g \circ g)(1) = 1$
5. $(f \circ g)(1.5) = -0.5$

Do you want details or have any questions?

Related Questions:

1. What is the derivative of $(f \circ g)(x)$?
2. What is the inverse of $f(x)$?
3. Find $(g \circ f)(-2)$.
4. Calculate $(f \circ f)(2)$.
5. Find the critical points of $(g \circ g)(x)$.
6. What is $(g \circ f)(3)$?
7. Determine the range of $(f \circ g)(x)$.
8. Calculate $(f \circ g)(0)$.

Tip:

When solving composite functions, always start by computing the inner function first and then substitute its result into the outer function.