The solution set of the inequality `(tan^(-1)x)^(2) le (tan^(-1)x) +6` is

Options

A. `[-tan2, tan3]`
B. `[tan2, - tan3]`
C. `[tan2, tan3]`
D. `(-oo, oo)`

Answer

To solve the inequality \((\tan^{-1} x)^2 \leq \tan^{-1} x + 6\), we will follow these steps:

### Step 1: Substitute \(\tan^{-1} x\) with \(t\)

Let \(t = \tan^{-1} x\). The inequality then becomes:
\[ t^2 \leq t + 6 \]

### Step 2: Rearrange the inequality

Rearranging gives us:
\[ t^2 - t - 6 \leq 0 \]

### Step 3: Factor the quadratic expression

We need to factor \(t^2 - t - 6\). We can look for two numbers that multiply to \(-6\) and add to \(-1\). These numbers are \(-3\) and \(2\). Thus, we can factor the expression as:
\[ (t - 3)(t + 2) \leq 0 \]

### Step 4: Find the critical points

The critical points from the factors are:
\[ t - 3 = 0 \implies t = 3 \]
\[ t + 2 = 0 \implies t = -2 \]

### Step 5: Test intervals around the critical points

We will test the intervals determined by the critical points \(-2\) and \(3\):
1. **Interval 1:** \(t < -2\) (e.g., \(t = -3\))
- \((t - 3)(t + 2) = (-3 - 3)(-3 + 2) = (-6)(-1) > 0\)

2. **Interval 2:** \(-2 < t < 3\) (e.g., \(t = 0\))
- \((t - 3)(t + 2) = (0 - 3)(0 + 2) = (-3)(2) < 0\)

3. **Interval 3:** \(t > 3\) (e.g., \(t = 4\))
- \((t - 3)(t + 2) = (4 - 3)(4 + 2) = (1)(6) > 0\)

### Step 6: Determine the solution set

The inequality \((t - 3)(t + 2) \leq 0\) holds true in the interval:
\[ [-2, 3] \]

### Step 7: Substitute back to find \(x\)

Now we substitute back \(t = \tan^{-1} x\):
\[ -2 \leq \tan^{-1} x \leq 3 \]

### Step 8: Find the values of \(x\)

To find the values of \(x\), we take the tangent of the endpoints:
1. \(\tan(-2)\)
2. \(\tan(3)\)

Thus, the solution set for \(x\) is:
\[ [\tan(-2), \tan(3)] \]

### Final Answer

The solution set of the inequality \((\tan^{-1} x)^2 \leq \tan^{-1} x + 6\) is:
\[ x \in [\tan(-2), \tan(3)] \]

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