Let's solve the given problem step by step.
### a. Finding [tex]\((f \circ g)(x)\)[/tex]
The function [tex]\((f \circ g)(x)\)[/tex] means [tex]\(f(g(x))\)[/tex]. To find it, we need to plug [tex]\(g(x)\)[/tex] into the function [tex]\(f(x)\)[/tex].
Recall that:
- [tex]\(f(x) = x - 3\)[/tex]
- [tex]\(g(x) = 5x^2 - 3\)[/tex]
So, we start with [tex]\(g(x)\)[/tex]:
[tex]\[g(x) = 5x^2 - 3\][/tex]
Now, substituting [tex]\(g(x)\)[/tex] into [tex]\(f\)[/tex]:
[tex]\[f(g(x)) = f(5x^2 - 3) = (5x^2 - 3) - 3 = 5x^2 - 6\][/tex]
Hence:
[tex]\[(f \circ g)(x) = 5x^2 - 6\][/tex]
### b. Finding [tex]\((g \circ f)(x)\)[/tex]
The function [tex]\((g \circ f)(x)\)[/tex] means [tex]\(g(f(x))\)[/tex]. To find it, we need to plug [tex]\(f(x)\)[/tex] into the function [tex]\(g(x)\)[/tex].
Recall that:
- [tex]\(f(x) = x - 3\)[/tex]
- [tex]\(g(x) = 5x^2 - 3\)[/tex]
So, we start with [tex]\(f(x)\)[/tex]:
[tex]\[f(x) = x - 3\][/tex]
Now, substituting [tex]\(f(x)\)[/tex] into [tex]\(g\)[/tex]:
[tex]\[g(f(x)) = g(x - 3) = 5(x - 3)^2 - 3\][/tex]
We need to expand [tex]\(5(x - 3)^2 - 3\)[/tex]:
[tex]\[(x - 3)^2 = x^2 - 6x + 9\][/tex]
Therefore:
[tex]\[g(f(x)) = 5(x^2 - 6x + 9) - 3 = 5x^2 - 30x + 45 - 3 = 5x^2 - 30x + 42\][/tex]
Hence:
[tex]\[(g \circ f)(x) = 5x^2 - 30x + 42\][/tex]
### c. Finding [tex]\((f \circ g)(1)\)[/tex]
To find [tex]\((f \circ g)(1)\)[/tex], we need to substitute [tex]\(x = 1\)[/tex] into the function [tex]\((f \circ g)(x)\)[/tex].
Recall:
[tex]\[(f \circ g)(x) = 5x^2 - 6\][/tex]
So:
[tex]\[(f \circ g)(1) = 5(1)^2 - 6 = 5 - 6 = -1\][/tex]
Hence:
[tex]\[(f \circ g)(1) = -1\][/tex]
### d. Finding [tex]\((g \circ f)(1)\)[/tex]
To find [tex]\((g \circ f)(1)\)[/tex], we need to substitute [tex]\(x = 1\)[/tex] into the function [tex]\((g \circ f)(x)\)[/tex].
Recall:
[tex]\[(g \circ f)(x) = 5x^2 - 30x + 42\][/tex]
So:
[tex]\[(g \circ f)(1) = 5(1)^2 - 30(1) + 42 = 5 - 30 + 42 = 17\][/tex]
Hence:
[tex]\[(g \circ f)(1) = 17\][/tex]
### Summary
- [tex]\((f \circ g)(x) = 5x^2 - 6\)[/tex]
- [tex]\((g \circ f)(x) = 5x^2 - 30x + 42\)[/tex]
- [tex]\((f \circ g)(1) = -1\)[/tex]
- [tex]\((g \circ f)(1) = 17\)[/tex]
