It is given that function f(x) = x² + 2x – 5

f’(x) = 2x + 2

If f’(x) = 0, then we get,

⇒ x = -1

So, the point x = -1 divides the real line into two disjoint intervals, (-∞,-1) and (1,∞)

So, in interval (-∞,-1)

f’(x) = 2x + 2 < 0

Therefore, the given function (f) is strictly decreasing in interval (-∞,-1).

And in interval (1,∞)

f’(x) = 2x + 2 > 0

Therefore, the given function (f) is strictly increasing in interval (1,∞).

Thus, f is strictly increasing for x > -1.