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Evaluate the following composition of functions:

Given:
[tex]\[ f(x) = 2x + 1, \quad g(x) = 5x^2, \quad h(x) = x + 3 \][/tex]

1. [tex]\((f \circ g)(x)\)[/tex]
2. [tex]\((g \circ f)(x)\)[/tex]
3. [tex]\((h \circ g)(x)\)[/tex]
4. [tex]\((f \circ h)(x)\)[/tex]


Sagot :

Certainly! Let's evaluate the given compositions of functions step-by-step.

Given functions:
[tex]\[ f(x) = 2x + 1 \][/tex]
[tex]\[ g(x) = 5x^2 \][/tex]
[tex]\[ h(x) = x + 3 \][/tex]

1. Evaluate [tex]\((f \circ g)(x)\)[/tex]:
[tex]\[ (f \circ g)(x) = f(g(x)) \][/tex]
First, we find [tex]\(g(x)\)[/tex]:
[tex]\[ g(x) = 5x^2 \][/tex]
Now, substitute [tex]\(g(x)\)[/tex] into [tex]\(f\)[/tex]:
[tex]\[ f(g(x)) = f(5x^2) = 2(5x^2) + 1 = 10x^2 + 1 \][/tex]
So, [tex]\((f \circ g)(x) = 10x^2 + 1\)[/tex].

For [tex]\(x = 2\)[/tex]:
[tex]\[ (f \circ g)(2) = 10(2^2) + 1 = 10 \cdot 4 + 1 = 40 + 1 = 41 \][/tex]

2. Evaluate [tex]\((g \circ f)(x)\)[/tex]:
[tex]\[ (g \circ f)(x) = g(f(x)) \][/tex]
First, we find [tex]\(f(x)\)[/tex]:
[tex]\[ f(x) = 2x + 1 \][/tex]
Now, substitute [tex]\(f(x)\)[/tex] into [tex]\(g\)[/tex]:
[tex]\[ g(f(x)) = g(2x + 1) = 5(2x + 1)^2 \][/tex]

So, [tex]\((g \circ f)(x)\)[/tex]:
For [tex]\(x = 2\)[/tex]:
[tex]\[ (g \circ f)(2) = 5((2 \cdot 2) + 1)^2 = 5(4 + 1)^2 = 5(5)^2 = 5 \cdot 25 = 125 \][/tex]

3. Evaluate [tex]\((h \circ g)(x)\)[/tex]:
[tex]\[ (h \circ g)(x) = h(g(x)) \][/tex]
First, we find [tex]\(g(x)\)[/tex]:
[tex]\[ g(x) = 5x^2 \][/tex]
Now, substitute [tex]\(g(x)\)[/tex] into [tex]\(h\)[/tex]:
[tex]\[ h(g(x)) = h(5x^2) = 5x^2 + 3 \][/tex]

So, [tex]\((h \circ g)(x) = 5x^2 + 3\)[/tex].

For [tex]\(x = 2\)[/tex]:
[tex]\[ (h \circ g)(2) = 5(2^2) + 3 = 5 \cdot 4 + 3 = 20 + 3 = 23 \][/tex]

4. Evaluate [tex]\((f \circ h)(x)\)[/tex]:
[tex]\[ (f \circ h)(x) = f(h(x)) \][/tex]
First, we find [tex]\(h(x)\)[/tex]:
[tex]\[ h(x) = x + 3 \][/tex]
Now, substitute [tex]\(h(x)\)[/tex] into [tex]\(f\)[/tex]:
[tex]\[ f(h(x)) = f(x + 3) = 2(x + 3) + 1 = 2x + 6 + 1 = 2x + 7 \][/tex]

So, [tex]\((f \circ h)(x) = 2x + 7\)[/tex].

For [tex]\(x = 2\)[/tex]:
[tex]\[ (f \circ h)(2) = 2(2 + 3) + 1 = 2 \cdot 2 + 7 = 4 + 7 = 11 \][/tex]

In summary, the evaluated compositions for [tex]\(x = 2\)[/tex] are:
1. [tex]\((f \circ g)(2) = 41\)[/tex]
2. [tex]\((g \circ f)(2) = 125\)[/tex]
3. [tex]\((h \circ g)(2) = 23\)[/tex]
4. [tex]\((f \circ h)(2) = 11\)[/tex]