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

Answer: 2 roots. The equation e^x + 0.5x^2 - 2 = 0 has exactly two solutions in the interval [-5, 5].

Step 1 - Set up the function. Let f(x) = e^x + 0.5x^2 - 2. The roots of the equation are exactly the points where f(x) = 0.

Step 2 - Bound the number of roots using convexity. Differentiating twice, f'(x) = e^x + x and f''(x) = e^x + 1. Since e^x > 0 for all real x, we have f''(x) = e^x + 1 > 0 everywhere, so f is strictly convex on the whole real line. A strictly convex function is shaped like a single valley, so its graph can cross the x-axis at most twice. Hence f has at most two roots.

Step 3 - Locate the minimum. Setting f'(x) = e^x + x = 0 gives a single critical point near x = -0.567 (since f' is itself increasing, this is the unique minimum). There f(x) is about -1.27, which is negative, so the lowest point of the curve lies below the x-axis.

Step 4 - Check the endpoints and confirm two crossings. Evaluate: f(-5) = e^{-5} + 12.5 - 2 is about 10.51 > 0, the minimum value is about -1.27 < 0, and f(5) = e^5 + 12.5 - 2 is about 158.91 > 0. The curve starts above the axis at x = -5, dips below the axis around its minimum, and rises back above the axis by x = 5.

Step 5 - Conclude. By the Intermediate Value Theorem there is one root on the way down (between x = -5 and the minimum, approximately x = -1.93) and one root on the way up (between the minimum and x = 5, approximately x = 0.60). Combined with the convexity bound of at most two roots, the equation has exactly two roots in [-5, 5].