The equation at hand is x12 + 4x – 1 = 0. To find the solutions, we can analyze the function defined by the left-hand side of the equation. Let’s define:
f(x) = x12 + 4x – 1
We want to determine the number of x-values for which f(x) = 0. First, we should investigate the behavior of this function as x approaches different limits:
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As x approaches positive infinity (x → ∞), f(x) clearly approaches positive infinity since the x12 term dominates and grows without bound.
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As x approaches negative infinity (x → -∞), f(x) also approaches positive infinity because the even-power term (x12) remains positive.
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At x = 0, we find f(0) = 012 + 4(0) – 1 = -1.
This immediate analysis implies that the function has at least one root because it must cross the x-axis, moving from negative (at x = 0) to positive (as x → ∞). To narrow it down, we can check the derivative:
f'(x) = 12x11 + 4.
Since 12x11 is non-negative (≥0) for all x and 4 is always positive, f'(x) is always greater than zero. Hence, f(x) is a strictly increasing function.
Being strictly increasing means that the function can only cross the x-axis once. Therefore, we conclude that there is exactly one real solution to the equation x12 + 4x – 1 = 0.