**Problem**

The equation \(x^3+3x-4=0\) has exactly one real root.

Discussion

*Hint: *Try differentiating. Is the function monotonic?

Note that cubic equation has at least one real root (fundamental theorem of algebra provides for three roots counting repetitions, real or complex; for equations with real coefficients, the complex roots appear in conjugate pairs. Hence a cubic equation with real coefficients must have at least one real root).

Now differentiating \( P(x) = x^3 + 3x – 4 \) we have the following: $$ \frac {d} {dx} P(x) = 3x^2 + 3 $$

What can you say about this derivative?

Clearly it is always positive (perfect square plus three is at least three or more).

Hence P(x) is monotonically increasing. It will ‘cut’ the x-axis only at one point.

Therefore P(x) has exactly one real root.

Remark:

In this problem, first we established P(x) has **at least **one real root. Then we proved it has **exactly one. ** Both steps are crucial.

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