Number of real roots of the equation `3x^(5) + 15x -8=0` is

Answer

To determine the number of real roots of the equation \(3x^5 + 15x - 8 = 0\), we will analyze the function and its derivatives step by step.

### Step 1: Define the function
Let \(f(x) = 3x^5 + 15x - 8\).

### Step 2: Find the first derivative
We need to find the first derivative \(f'(x)\) to analyze the behavior of the function.
\[ f'(x) = \frac{d}{dx}(3x^5 + 15x - 8) = 15x^4 + 15 \]

### Step 3: Analyze the first derivative
Notice that \(f'(x) = 15x^4 + 15\). Since \(15x^4\) is always non-negative (as any even power is non-negative) and \(15\) is positive, we have:
\[ f'(x) > 0 \quad \text{for all } x \]
This means that the function \(f(x)\) is always increasing.

### Step 4: Evaluate the function at a specific point
To find the number of real roots, we can evaluate \(f(x)\) at \(x = 0\):
\[ f(0) = 3(0)^5 + 15(0) - 8 = -8 \]
Since \(f(0) = -8 < 0\), we know that the function is below the x-axis at \(x = 0\).

### Step 5: Check the behavior as \(x\) approaches infinity
As \(x\) approaches positive infinity, \(f(x)\) will also approach positive infinity because the term \(3x^5\) dominates:
\[ \lim_{x \to +\infty} f(x) = +\infty \]

### Step 6: Conclusion on the number of roots
Since \(f(x)\) is continuous, always increasing, and changes from negative at \(x = 0\) to positive as \(x\) approaches infinity, there must be exactly one point where \(f(x) = 0\). Therefore, the equation \(3x^5 + 15x - 8 = 0\) has exactly one real root.

### Final Answer
The number of real roots of the equation \(3x^5 + 15x - 8 = 0\) is **1**.

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