The number of solution of equation `x^3-21 x-20=e^x` is/are

Options

A. `0`
B. `1`
C. `2`
D. `3`

Answer

To find the number of solutions of the equation \( x^3 - 21x - 20 = e^x \), we can follow these steps:

### Step 1: Define the Functions
Let \( f(x) = x^3 - 21x - 20 \) and \( g(x) = e^x \). We need to find the points where \( f(x) = g(x) \).

### Step 2: Analyze the Function \( f(x) \)
To understand the behavior of \( f(x) \), we can find its critical points by calculating the derivative:
\[ f'(x) = 3x^2 - 21 \]
Set the derivative to zero to find critical points:
\[ 3x^2 - 21 = 0 \implies x^2 = 7 \implies x = \pm \sqrt{7} \]

### Step 3: Evaluate \( f(x) \) at Critical Points
Calculate \( f(x) \) at the critical points:
\[ f(\sqrt{7}) = (\sqrt{7})^3 - 21(\sqrt{7}) - 20 \]
\[ f(-\sqrt{7}) = (-\sqrt{7})^3 - 21(-\sqrt{7}) - 20 \]

### Step 4: Determine the Behavior of \( f(x) \)
- As \( x \to -\infty \), \( f(x) \to -\infty \).
- As \( x \to +\infty \), \( f(x) \to +\infty \).
- Check the values of \( f(x) \) at the critical points to find local maxima and minima.

### Step 5: Analyze the Function \( g(x) \)
The function \( g(x) = e^x \) is always positive and increases exponentially.

### Step 6: Find Intersections
Since \( f(x) \) is a cubic polynomial and \( g(x) \) is an exponential function, we can analyze their graphs:
- The cubic function \( f(x) \) will have at most 3 real roots (due to its degree).
- The exponential function \( g(x) \) will intersect the cubic function at points where \( f(x) = g(x) \).

### Step 7: Conclusion
By analyzing the graphs of \( f(x) \) and \( g(x) \), we find that they intersect at three points. Therefore, the number of solutions to the equation \( x^3 - 21x - 20 = e^x \) is:

\[ \text{Number of solutions} = 3 \]

### Final Answer
The number of solutions of the equation \( x^3 - 21x - 20 = e^x \) is **3**.

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