To solve the equation [tex]\(\log(x+1) = -x^2 + 10\)[/tex] graphically, let's break this down step-by-step.
### Step 1: Understanding the equations
We are dealing with two functions:
1. [tex]\(y_1 = \log(x + 1)\)[/tex]
2. [tex]\(y_2 = -x^2 + 10\)[/tex]
### Step 2: Sketching the functions
1. Plotting [tex]\(y_1 = \log(x + 1)\)[/tex]:
- The function [tex]\(\log(x + 1)\)[/tex] is defined only for [tex]\(x > -1\)[/tex] because the argument of the logarithm must be positive.
- As [tex]\(x\)[/tex] approaches [tex]\(-1\)[/tex] from the right, [tex]\(\log(x + 1)\)[/tex] approaches [tex]\(-\infty\)[/tex].
- As [tex]\(x\)[/tex] increases, [tex]\(\log(x + 1)\)[/tex] increases at a decreasing rate.
2. Plotting [tex]\(y_2 = -x^2 + 10\)[/tex]:
- This is a downward-opening parabola with its vertex at [tex]\( (0, 10) \)[/tex].
- The function crosses the y-axis at (0, 10) and is symmetric about the y-axis.
- It intersects the x-axis where [tex]\(x^2 = 10\)[/tex], i.e., at [tex]\(x = \pm \sqrt{10} \approx \pm 3.16\)[/tex].
### Step 3: Finding the intersection points
To find the approximate solution, we need to determine where these two functions intersect. Let's analyze this step-by-step:
1. Evaluating [tex]\(y_1 = \log(x + 1)\)[/tex]:
- For positive values of [tex]\(x\)[/tex], [tex]\(y_1\)[/tex] increases slowly.
- For small positive values ([tex]\(x > 0\)[/tex]), [tex]\(\log(x + 1)\)[/tex] starts slightly above 0 and increases gradually.
- For large values of [tex]\(x\)[/tex], it continues to increase but at a slower rate.
2. Evaluating [tex]\(y_2 = -x^2 + 10\)[/tex]:
- For values near [tex]\(x = 0\)[/tex], [tex]\(y_2\)[/tex] starts at 10 and decreases rapidly.
- At [tex]\(x = \pm 1\)[/tex], [tex]\(y_2 = 9\)[/tex].
- At [tex]\(x = \pm 2\)[/tex], [tex]\(y_2 = 6\)[/tex].
- At [tex]\(x = \pm 3\)[/tex], [tex]\(y_2 = 1\)[/tex].
### Step 4: Intersect graphically
Since we are approximating the solution graphically:
1. Start from the positive [tex]\(x\)[/tex]-values. We identify the logarithmic function intersecting the parabola.
2. Around [tex]\(x \approx 2\)[/tex], [tex]\(\log(2+1) \approx \log(3) \approx 1.1\)[/tex] and, for the parabola, [tex]\((-2^2 + 10 = 6)\)[/tex].
On further inspection, we see:
- Between [tex]\(x = 2\)[/tex] and [tex]\(x = 2.5\)[/tex], should yield a value closer to where [tex]\(\log(x+1) \approx -x^2 + 10\)[/tex].
### Step 5: Numerical check
Evaluate near [tex]\(x = 2.2\)[/tex]:
- [tex]\(\log(2.2+1) \approx \log(3.2) \approx 1.16\)[/tex]
- [tex]\(-2.2^2 + 10 \approx -4.84 + 10 \approx 5.16\)[/tex]
[tex]\(\log(2.5+1) \approx \log(3.5) \approx 1.25\)[/tex]
[tex]\(-2.5^2 + 10 \approx -6.25 + 10 \approx 3.75\)[/tex]
Exact intersection requires more precise solving or graphing software, approximating \( x \approx +2.5).
Beyond analytical approach, relying above evaluation closely, graphical software exact may help verifying closer intersection points around these values.
### Conclusion
By visually and analytically intersection on above equations:
Approximate solution:
[tex]\[ x \approx 2.2 - 2.3. \][/tex]
This concludes our process for finding solutions graphically and numerically evaluating intersection points.
