Composite Functions - Examples

Question 1: Computing Composite Functions (from file image_c62638.png)

The Question: If $f(x) = 10(x-5)^2 + 100x - 225$ and $g(x) = \sqrt{2x+5}$, then the composite functions $f(g(x))$ and $g(f(x))$ are respectively __________.

Detailed Solution:

First, let’s simplify the function $f(x)$ to make the substitution easier.

• $f(x) = 10(x^2 - 10x + 25) + 100x - 225$
• $f(x) = 10x^2 - 100x + 250 + 100x - 225$
• The $-100x$ and $+100x$ terms cancel out.
• $f(x) = 10x^2 + 25$

Now we can compute the compositions.

1. Find $f(g(x))$:

• This means we substitute the entire expression for $g(x)$ into the $x$ of our simplified $f(x)$.
• $f(g(x)) = 10(g(x))^2 + 25$
• $f(g(x)) = 10(\sqrt{2x+5})^2 + 25$
• The square and the square root cancel each other out:
• $f(g(x)) = 10(2x+5) + 25$
• $f(g(x)) = 20x + 50 + 25 = 20x + 75$

2. Find $g(f(x))$:

• This means we substitute the entire expression for $f(x)$ into the $x$ of $g(x)$.
• $g(f(x)) = \sqrt{2(f(x)) + 5}$
• $g(f(x)) = \sqrt{2(10x^2 + 25) + 5}$
• $g(f(x)) = \sqrt{20x^2 + 50 + 5}$
• $g(f(x)) = \sqrt{20x^2 + 55}$

Final Answer: The composite functions are $20x + 75$ and $\sqrt{20x^2 + 55}$.

Question 2: Evaluating a Composite Function (from file image_c62638.png)

The Question: Suppose $f(x) = 3x+10$ and $g(x) = \sqrt{x+11}$ are two functions. Find the value of $f(g(5))$.

Detailed Solution:

We evaluate this from the inside out.

1. First, calculate the inner function, $g(5)$:
   • $g(5) = \sqrt{5+11} = \sqrt{16} = 4$.
2. Next, use this result as the input for the outer function, $f(x)$:
   • We need to find $f(4)$.
   • $f(4) = 3(4) + 10 = 12 + 10 = 22$.

Final Answer: The value of $f(g(5))$ is 22.

Question 3: Evaluating a Composite Function (Reversed) (from file image_c62638.png)

The Question: Suppose $f(x) = 3x+10$ and $g(x) = \sqrt{x+11}$ are two functions. Find the value of $g(f(5))$.

Detailed Solution:

This is similar to the last question, but the order of the functions is reversed.

1. First, calculate the inner function, $f(5)$:
   • $f(5) = 3(5) + 10 = 15 + 10 = 25$.
2. Next, use this result as the input for the outer function, $g(x)$:
   • We need to find $g(25)$.
   • $g(25) = \sqrt{25+11} = \sqrt{36} = 6$.

Final Answer: The value of $g(f(5))$ is 6.

Question 4: Checking for Commutativity (from file image_c62638.png)

The Question: Which of the following pairs of functions $f(x)$ and $g(x)$ satisfies the equation $f(g(x)) = g(f(x))$?

Detailed Solution:

We must compute both compositions for each pair and check if the results are identical.

• Pair 1: $f(x) = 2x$ and $g(x) = 10x+3$

  • $f(g(x)) = 2(10x+3) = 20x+6$
  • $g(f(x)) = 10(2x)+3 = 20x+3$. (Not equal)

• Pair 2: $f(x) = 7x+6$ and $g(x) = 4x+3$

  • $f(g(x)) = 7(4x+3)+6 = 28x+21+6 = 28x+27$
  • $g(f(x)) = 4(7x+6)+3 = 28x+24+3 = 28x+27$. (Equal)

• Pair 3: $f(x) = e^x$ and $g(x) = x$

  • $f(g(x)) = e^{(x)} = e^x$
  • $g(f(x)) = (e^x) = e^x$. (Equal)

• Pair 4: $f(x) = 4e^x$ and $g(x) = 4x$

  • $f(g(x)) = 4e^{(4x)} = 4e^{4x}$
  • $g(f(x)) = 4(4e^x) = 16e^x$. (Not equal)

Final Answer: The pairs that satisfy the equation are:

• $f(x) = 7x+6$ and $g(x) = 4x+3$
• $f(x) = e^x$ and $g(x) = x$

Question 5: Comparing Function Values (from file image_c6232e.png)

The Question: If $f(x) = 2\sqrt{x} + 9$, $g(x) = 8x^2$, and $h(x) = 5x+7$, then which of the following options are CORRECT?

Detailed Solution:

Let’s evaluate each statement carefully.

• $f(g(0)) < g(f(0))$

  • First, $g(0) = 8(0)^2 = 0$. So $f(g(0)) = f(0) = 2\sqrt{0}+9 = 9$.
  • Second, $f(0) = 9$. So $g(f(0)) = g(9) = 8(9)^2 = 8(81) = 648$.
  • Is $9 < 648$? Yes. This statement is TRUE.

• $h(f(0)) > f(h(0))$

  • First, $f(0) = 9$. So $h(f(0)) = h(9) = 5(9)+7 = 45+7 = 52$.
  • Second, $h(0) = 5(0)+7 = 7$. So $f(h(0)) = f(7) = 2\sqrt{7}+9$. Since $\sqrt{4}=2$ and $\sqrt{9}=3$, $\sqrt{7}$ is about 2.6. So $f(7) \approx 2(2.6)+9 = 5.2+9 = 14.2$.
  • Is $52 > 14.2$? Yes. This statement is TRUE.

• $f(g(x)) < g(f(x))$, for all $x \in (-\infty, +\infty)$

  • The domain of $f(x)$ is $x \ge 0$. The composite function $g(f(x)) = 8(2\sqrt{x}+9)^2$ is also only defined for $x \ge 0$. Since the statement requires the inequality to hold for all real numbers (including negatives), it cannot be correct. This statement is FALSE.

• $f(h(x)) < h(f(x))$, for all $x \in [0, +\infty)$

  • Let’s find the functions:
    • $h(f(x)) = 5(f(x))+7 = 5(2\sqrt{x}+9)+7 = 10\sqrt{x}+45+7 = 10\sqrt{x}+52$.
    • $f(h(x)) = 2\sqrt{h(x)}+9 = 2\sqrt{5x+7}+9$.
  • Let’s test a large value, $x=100$.
    • $h(f(100)) = 10\sqrt{100}+52 = 10(10)+52 = 152$.
    • $f(h(100)) = 2\sqrt{5(100)+7}+9 = 2\sqrt{507}+9 \approx 2(22.5)+9 = 54$.
  • For $x=100$, $54 < 152$. The linear term inside h(f(x)) grows much faster than the square root term inside f(h(x)). This statement appears to be TRUE.

• $h(g(1)) > g(h(1))$

  • First, $g(1) = 8(1)^2=8$. So $h(g(1)) = h(8) = 5(8)+7 = 47$.
  • Second, $h(1) = 5(1)+7=12$. So $g(h(1)) = g(12) = 8(12)^2 = 8(144) = 1152$.
  • Is $47 > 1152$? No. This statement is FALSE.

Final Answer: The correct statements are:

• $f(g(0)) < g(f(0))$
• $h(f(0)) > f(h(0))$
• $f(h(x)) < h(f(x))$, for all $x \in [0, +\infty)$