24 Select the correct answer. Consider functions $$ f $$ and $$ g $$. $$ f(x)=x^{4}+9 x^{2}-3 $$ $$ g(x)=\left(\frac{1}{2}\right)^{x-2} $$ Using a table of values, what are the approximate solutions to the equation $$ f(x)=g(x) $$ to the nearest quarter of a unit? A. $$ \quad x \approx-0.25 $$ and $$ x \approx 2.5 $$ B. $$ \quad x \approx-1 $$ and $$ x \approx 0.75 $$ C. $$ \quad x \approx-0.5 $$ and $$ x \approx 0.5 $$ D. $$ \quad x \approx-0.25 $$ and $$ x \approx 1.5 $$ Reset Next

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To find the approximate solutions to the equation $$ f(x) = g(x) $$, we will evaluate both functions $$ f(x) $$ and $$ g(x) $$ at various values of $$ x $$ and look for points where they are approximately equal.

### Step 1: Define the functions
- $$ f(x) = x^4 + 9x^2 - 3 $$
- $$ g(x) = \left(\frac{1}{2}\right)^{x-2} $$

### Step 2: Create a table of values
We will calculate $$ f(x) $$ and $$ g(x) $$ for a range of $$ x $$ values to find where they intersect.

Let's evaluate $$ f(x) $$ and $$ g(x) $$ for $$ x $$ values from -2 to 3 in increments of 0.25.

### Step 3: Calculate values

1. **For $$ x = -2 $$**:
   - $$ f(-2) = (-2)^4 + 9(-2)^2 - 3 = 16 + 36 - 3 = 49 $$
   - $$ g(-2) = \left(\frac{1}{2}\right)^{-2-2} = \left(\frac{1}{2}\right)^{-4} = 16 $$

2. **For $$ x = -1.75 $$**:
   - $$ f(-1.75) = (-1.75)^4 + 9(-1.75)^2 - 3 \approx 29.64 $$
   - $$ g(-1.75) = \left(\frac{1}{2}\right)^{-1.75-2} \approx 5.66 $$

3. **For $$ x = -1.5 $$**:
   - $$ f(-1.5) = (-1.5)^4 + 9(-1.5)^2 - 3 \approx 18.69 $$
   - $$ g(-1.5) = \left(\frac{1}{2}\right)^{-1.5-2} \approx 3.17 $$

4. **For $$ x = -1.25 $$**:
   - $$ f(-1.25) = (-1.25)^4 + 9(-1.25)^2 - 3 \approx 10.39 $$
   - $$ g(-1.25) = \left(\frac{1}{2}\right)^{-1.25-2} \approx 2.00 $$

5. **For $$ x = -1 $$**:
   - $$ f(-1) = (-1)^4 + 9(-1)^2 - 3 = 1 + 9 - 3 = 7 $$
   - $$ g(-1) = \left(\frac{1}{2}\right)^{-1-2} = \left(\frac{1}{2}\right)^{-3} = 8 $$

6. **For $$ x = -0.75 $$**:
   - $$ f(-0.75) = (-0.75)^4 + 9(-0.75)^2 - 3 \approx 2.39 $$
   - $$ g(-0.75) = \left(\frac{1}{2}\right)^{-0.75-2} \approx 4.00 $$

7. **For $$ x = -0.5 $$**:
   - $$ f(-0.5) = (-0.5)^4 + 9(-0.5)^2 - 3 \approx -0.44 $$
   - $$ g(-0.5) = \left(\frac{1}{2}\right)^{-0.5-2} \approx 2.83 $$

8. **For $$ x = 0 $$**:
   - $$ f(0) = 0^4 + 9(0)^2 - 3 = -3 $$
   - $$ g(0) = \left(\frac{1}{2}\right)^{0-2} = \frac{1}{4} = 0.25 $$

9. **For $$ x = 0.5 $$**:
   - $$ f(0.5) = (0.5)^4 + 9(0.5)^2 - 3 \approx -1.44 $$
   - $$ g(0.5) = \left(\frac{1}{2}\right)^{0.5-2} \approx 1.41 $$

10. **For $$ x = 1 $$**:
    - $$ f(1) = 1^4 + 9(1)^2 - 3 = 1 + 9 - 3 = 7 $$
    - $$ g(1) = \left(\frac{1}{2}\right)^{1-2} = 2 $$

11. **For $$ x = 1.5 $$**:
    - $$ f(1.5) = (1.5)^4 + 9(1.5)^2 - 3 \approx 18.69 $$
    - $$ g(1.5) = \left(\frac{1}{2}\right)^{1.5-2} \approx 1.41 $$

12. **For $$ x = 2 $$**:
    - $$ f(2) = 2^4 + 9(2)^2 - 3 = 16 + 36 - 3 = 49 $$
    - $$ g(2) = \left(\frac{1}{2}\right)^{2-2} = 1 $$

13. **For $$ x = 2.5 $$**:
    - $$ f(2.5) = (2.5)^4 + 9(2.5)^2 - 3 \approx 90.06 $$
    - $$ g(2.5) = \left(\frac{1}{2}\right)^{2.5-2} \approx 0.707 $$

### Step 4: Analyze the results
From the calculations, we can see that:

- At $$ x \approx -1 $$, $$ f(-1) \approx 7 $$ and $$ g(-1) \approx 8 $$ (close).
- At $$ x \approx -0.75 $$, $$ f(-0.75) \approx 2.39 $$ and $$ g(-0.75) \approx 4.00 $$ (getting closer).
- At $$ x \approx 0.5 $$, $$ f(0.5) \approx -1.44 $$ and $$ g(0.5) \approx 1.41 $$ (crossing zero).
- At $$ x \approx 1 $$, $$ f(1) \approx 7 $$ and $$ g(1) \approx 2 $$ (still diverging).

### Conclusion
The approximate solutions to the equation $$ f(x) = g(x
The approximate solutions are $$ x \approx -0.25 $$ and $$ x \approx 2.5 $$.

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