The number of roots of \(e^{x}+0.5x^{2}-2=0\) in the range \([-5,5]\) is

For a twice-differentiable function f, if f″(x) > 0 for every x in an interval then f is strictly convex there — its slope f′(x) is strictly increasing, so f can only decrease and then increase, never reverse a second time. Such a curve is shaped like a single valley, so it crosses the x-axis at most twice: once on the way down and once on the way back up, and only that many times if the valley floor actually dips below the axis while both ends stay above it.

Applying this to the given equation:

1. Let f(x) = ex + 0.5x2 − 2, so the roots of the equation are exactly the zeros of f.

2. Differentiate twice: f′(x) = ex + x and f″(x) = ex + 1. Since ex > 0 for every real x, f″(x) = ex + 1 > 0 everywhere, so f is strictly convex on all of ℝ, matching the concept above — it can have at most two real roots.

3. Evaluate at the interval's endpoints: f(-5) = e-5 + 12.5 − 2 ≈ 10.51 > 0 and f(5) = e5 + 12.5 − 2 ≈ 158.91 > 0 — both ends of [−5, 5] lie above the axis.

4. Evaluate at an interior point: f(0) = 1 + 0 − 2 = −1 < 0, so the curve dips below the axis somewhere inside the interval.

5. By the Intermediate Value Theorem, f changes sign once between −5 and 0 (positive to negative) and once between 0 and 5 (negative to positive) — two sign changes, i.e. two crossings, exactly matching the convexity bound of at most two roots.

Cross-check via the minimum: solving f′(x) = ex + x = 0 locates the single critical point (f′ is itself strictly increasing since f″ > 0, so it has only one zero) at x ≈ −0.567, where f(x) ≈ −1.27. This confirms a single valley floor below the axis, consistent with exactly two crossings rather than zero or more than two.

Hence the equation ex + 0.5x2 − 2 = 0 has exactly 2 roots in [−5, 5].