To find the restricted values in the expression \( \frac{4x}{x^2 + 16} \), we need to determine when the denominator is equal to zero since division by zero is undefined.
The denominator is \( x^2 + 16 \). Set it equal to zero:
\[ x^2 + 16 = 0 \]
Subtract 16 from both sides:
\[ x^2 = -16 \]
Now, take the square root of both sides. Since the square root of a negative number is not a real number, \( x^2 = -16 \) has no real solutions.
Thus, the denominator \( x^2 + 16 \) is never equal to zero for any real value of \( x \). This means that there are no restricted values for the expression \( \frac{4x}{x^2 + 16} \) in the set of real numbers.
In summary, the expression \( \frac{4x}{x^2 + 16} \) has no restricted values.
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