Math / Algebra

QuestionFor $f(x)=8 x-7$ and $g(x)=\frac{x+7}{8}$ , find the following functions. a. $(f \circ g)(x) ; b .(g \circ f)(x) ;$ c. $(f \circ g)(6) ; d .(g \circ f)(6)$ a. $(f \circ g)(x)=$ $\square$ (Simplify your answer.) b. $(g \circ f)(x)=$ $\square$ (Simplify your answer.) c. $(f \circ g)(6)=$ $\square$ d. $(g \circ f)(6)=$ $\square$

Studdy Solution

STEP 1

What is this asking? We're given two functions, $f(x)$ and $g(x)$ , and we need to find the composite functions $(f \circ g)(x)$ and $(g \circ f)(x)$ , as well as their values at $x = 6$ . Watch out! Remember the order matters in function composition! $(f \circ g)(x)$ means $f(g(x))$ , not $g(f(x))$ .

STEP 2

1. Find $(f \circ g)(x)$
2. Find $(g \circ f)(x)$
3. Find $(f \circ g)(6)$
4. Find $(g \circ f)(6)$

STEP 3

Let's start with $(f \circ g)(x)$ , which means we're putting $g(x)$ inside $f(x)$ .
It's like a function turducken!

STEP 4

We know $f(x) = 8x - 7$ and $g(x) = \frac{x + 7}{8}$ .
So, $(f \circ g)(x) = f(g(x))$ .
This means wherever we see an $x$ in $f(x)$ , we replace it with the entire $g(x)$ function.

STEP 5

$f(g(x)) = f\left(\frac{x + 7}{8}\right) = 8 \cdot \left(\frac{x + 7}{8}\right) - 7$

STEP 6

Now, we simplify: $8 \cdot \left(\frac{x + 7}{8}\right) - 7 = \frac{8}{8} \cdot (x+7) - 7 = 1 \cdot (x + 7) - 7 = x + 7 - 7 = x$ So, $(f \circ g)(x) = x$ !

STEP 7

Now, let's find $(g \circ f)(x)$ , which means $g(f(x))$ .
This time, $f(x)$ goes inside $g(x)$ .

STEP 8

We have $g(x) = \frac{x + 7}{8}$ and $f(x) = 8x - 7$ .
So, $g(f(x)) = g(8x - 7) = \frac{(8x - 7) + 7}{8}$

STEP 9

Let's simplify this: $\frac{8x - 7 + 7}{8} = \frac{8x}{8} = \frac{8}{8} \cdot x = 1 \cdot x = x$ So, $(g \circ f)(x) = x$ !

STEP 10

We already found that $(f \circ g)(x) = x$ .
So, to find $(f \circ g)(6)$ , we just substitute $x = 6$ : $(f \circ g)(6) = 6$

STEP 11

Similarly, we found $(g \circ f)(x) = x$ .
So, for $(g \circ f)(6)$ , we substitute $x = 6$ : $(g \circ f)(6) = 6$

STEP 12

a. $(f \circ g)(x) = x$ b. $(g \circ f)(x) = x$ c. $(f \circ g)(6) = 6$ d. $(g \circ f)(6) = 6$

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Studdy Solution

STEP 1

STEP 2

1. Find $(f \circ g)(x)$
2. Find $(g \circ f)(x)$
3. Find $(f \circ g)(6)$
4. Find $(g \circ f)(6)$

STEP 3

Let's start with $(f \circ g)(x)$ , which means we're putting $g(x)$ inside $f(x)$ .
It's like a function turducken!

STEP 4

We know $f(x) = 8x - 7$ and $g(x) = \frac{x + 7}{8}$ .
So, $(f \circ g)(x) = f(g(x))$ .
This means wherever we see an $x$ in $f(x)$ , we replace it with the entire $g(x)$ function.

STEP 5

$f(g(x)) = f\left(\frac{x + 7}{8}\right) = 8 \cdot \left(\frac{x + 7}{8}\right) - 7$

STEP 6

Now, we simplify: $8 \cdot \left(\frac{x + 7}{8}\right) - 7 = \frac{8}{8} \cdot (x+7) - 7 = 1 \cdot (x + 7) - 7 = x + 7 - 7 = x$ So, $(f \circ g)(x) = x$ !

STEP 7

Now, let's find $(g \circ f)(x)$ , which means $g(f(x))$ .
This time, $f(x)$ goes inside $g(x)$ .

STEP 8

We have $g(x) = \frac{x + 7}{8}$ and $f(x) = 8x - 7$ .
So, $g(f(x)) = g(8x - 7) = \frac{(8x - 7) + 7}{8}$

STEP 9

Let's simplify this: $\frac{8x - 7 + 7}{8} = \frac{8x}{8} = \frac{8}{8} \cdot x = 1 \cdot x = x$ So, $(g \circ f)(x) = x$ !

STEP 10

We already found that $(f \circ g)(x) = x$ .
So, to find $(f \circ g)(6)$ , we just substitute $x = 6$ : $(f \circ g)(6) = 6$

STEP 11

Similarly, we found $(g \circ f)(x) = x$ .
So, for $(g \circ f)(6)$ , we substitute $x = 6$ : $(g \circ f)(6) = 6$

STEP 12

a. $(f \circ g)(x) = x$ b. $(g \circ f)(x) = x$ c. $(f \circ g)(6) = 6$ d. $(g \circ f)(6) = 6$

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Studdy Solution

STEP 1

STEP 2

1. Find (f∘g)(x)(f \circ g)(x)(f∘g)(x)2. Find (g∘f)(x)(g \circ f)(x)(g∘f)(x)3. Find (f∘g)(6)(f \circ g)(6)(f∘g)(6)4. Find (g∘f)(6)(g \circ f)(6)(g∘f)(6)

STEP 3

Let's **start** with (f∘g)(x)(f \circ g)(x)(f∘g)(x), which means we're putting g(x)g(x)g(x) *inside* f(x)f(x)f(x).It's like a function turducken!

STEP 4

We know f(x)=8x−7f(x) = 8x - 7f(x)=8x−7 and g(x)=x+78g(x) = \frac{x + 7}{8}g(x)=8x+7​.So, (f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x))(f∘g)(x)=f(g(x)).This means wherever we see an xxx in f(x)f(x)f(x), we replace it with the *entire* g(x)g(x)g(x) function.

STEP 5

f(g(x))=f(x+78)=8⋅(x+78)−7 f(g(x)) = f\left(\frac{x + 7}{8}\right) = 8 \cdot \left(\frac{x + 7}{8}\right) - 7 f(g(x))=f(8x+7​)=8⋅(8x+7​)−7

STEP 6

Now, we **simplify**: 8⋅(x+78)−7=88⋅(x+7)−7=1⋅(x+7)−7=x+7−7=x 8 \cdot \left(\frac{x + 7}{8}\right) - 7 = \frac{8}{8} \cdot (x+7) - 7 = 1 \cdot (x + 7) - 7 = x + 7 - 7 = x 8⋅(8x+7​)−7=88​⋅(x+7)−7=1⋅(x+7)−7=x+7−7=x So, (f∘g)(x)=x(f \circ g)(x) = x(f∘g)(x)=x!

STEP 7

Now, let's find (g∘f)(x)(g \circ f)(x)(g∘f)(x), which means g(f(x))g(f(x))g(f(x)).This time, f(x)f(x)f(x) goes *inside* g(x)g(x)g(x).

STEP 8

We have g(x)=x+78g(x) = \frac{x + 7}{8}g(x)=8x+7​ and f(x)=8x−7f(x) = 8x - 7f(x)=8x−7.So, g(f(x))=g(8x−7)=(8x−7)+78 g(f(x)) = g(8x - 7) = \frac{(8x - 7) + 7}{8} g(f(x))=g(8x−7)=8(8x−7)+7​

STEP 9

Let's **simplify** this: 8x−7+78=8x8=88⋅x=1⋅x=x \frac{8x - 7 + 7}{8} = \frac{8x}{8} = \frac{8}{8} \cdot x = 1 \cdot x = x 88x−7+7​=88x​=88​⋅x=1⋅x=x So, (g∘f)(x)=x(g \circ f)(x) = x(g∘f)(x)=x!

STEP 10

We already found that (f∘g)(x)=x(f \circ g)(x) = x(f∘g)(x)=x.So, to find (f∘g)(6)(f \circ g)(6)(f∘g)(6), we just **substitute** x=6x = 6x=6: (f∘g)(6)=6 (f \circ g)(6) = 6 (f∘g)(6)=6

STEP 11

Similarly, we found (g∘f)(x)=x(g \circ f)(x) = x(g∘f)(x)=x.So, for (g∘f)(6)(g \circ f)(6)(g∘f)(6), we **substitute** x=6x = 6x=6: (g∘f)(6)=6 (g \circ f)(6) = 6 (g∘f)(6)=6

STEP 12

a. (f∘g)(x)=x(f \circ g)(x) = x(f∘g)(x)=x b. (g∘f)(x)=x(g \circ f)(x) = x(g∘f)(x)=x c. (f∘g)(6)=6(f \circ g)(6) = 6(f∘g)(6)=6 d. (g∘f)(6)=6(g \circ f)(6) = 6(g∘f)(6)=6

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1. Find $(f \circ g)(x)$
2. Find $(g \circ f)(x)$
3. Find $(f \circ g)(6)$
4. Find $(g \circ f)(6)$

Let's start with $(f \circ g)(x)$ , which means we're putting $g(x)$ inside $f(x)$ .
It's like a function turducken!

We know $f(x) = 8x - 7$ and $g(x) = \frac{x + 7}{8}$ .
So, $(f \circ g)(x) = f(g(x))$ .
This means wherever we see an $x$ in $f(x)$ , we replace it with the entire $g(x)$ function.

$f(g(x)) = f\left(\frac{x + 7}{8}\right) = 8 \cdot \left(\frac{x + 7}{8}\right) - 7$

Now, we simplify: $8 \cdot \left(\frac{x + 7}{8}\right) - 7 = \frac{8}{8} \cdot (x+7) - 7 = 1 \cdot (x + 7) - 7 = x + 7 - 7 = x$ So, $(f \circ g)(x) = x$ !

Now, let's find $(g \circ f)(x)$ , which means $g(f(x))$ .
This time, $f(x)$ goes inside $g(x)$ .

We have $g(x) = \frac{x + 7}{8}$ and $f(x) = 8x - 7$ .
So, $g(f(x)) = g(8x - 7) = \frac{(8x - 7) + 7}{8}$

Let's simplify this: $\frac{8x - 7 + 7}{8} = \frac{8x}{8} = \frac{8}{8} \cdot x = 1 \cdot x = x$ So, $(g \circ f)(x) = x$ !

We already found that $(f \circ g)(x) = x$ .
So, to find $(f \circ g)(6)$ , we just substitute $x = 6$ : $(f \circ g)(6) = 6$

Similarly, we found $(g \circ f)(x) = x$ .
So, for $(g \circ f)(6)$ , we substitute $x = 6$ : $(g \circ f)(6) = 6$

a. $(f \circ g)(x) = x$ b. $(g \circ f)(x) = x$ c. $(f \circ g)(6) = 6$ d. $(g \circ f)(6) = 6$