Level 7 · Ages 17–18
Read it again any time, including while you work through the exercises.
You already know the exponential function and its algebra rules. Time to paint its complete portrait, the way we study any function: sign, variations, limits, tangents. And this portrait is rarely elegant, because every property flows from one magic identity: the exponential function is its own derivative, (eˣ)′ = eˣ.
First property: for every real x, eˣ > 0. An exponential is never zero, never negative, even for very negative x: e⁻¹⁰⁰⁰ is tiny, but strictly positive. The curve therefore lives entirely above the x-axis.
Combine the two: the derivative of eˣ is eˣ, which is always strictly positive. A function whose derivative is strictly positive everywhere is strictly increasing on ℝ. A precious consequence for equations: eᵃ = eᵇ is equivalent to a = b, and eᵃ < eᵇ is equivalent to a < b.
At the far ends, the behaviors contrast sharply: as x tends to -∞, eˣ tends to 0 (the x-axis is an asymptote), and as x tends to +∞, eˣ tends to +∞, faster than any power of x. In between, the curve passes through two landmark points: (0, 1) since e⁰ = 1, and (1, e) with e ≈ 2.718.
Finally, the tangent at the point (0, 1) has slope e⁰ = 1: its equation is y = x + 1. The curve always stays above this tangent, a famous inequality you will prove in this lesson: eˣ ≥ x + 1 for every real x.
Find the equation of the tangent to the exp curve at the point with x-coordinate 0.
f(0) = e⁰ = 1
I compute the point of tangency: the curve passes through (0, 1).
f′(x) = eˣ donc f′(0) = 1
The exponential is its own derivative, so the slope of the tangent at 0 is e⁰ = 1.
y = f′(0)(x - 0) + f(0) = x + 1
I apply the tangent formula: slope 1, through (0, 1). I get y = x + 1.
Result: The tangent at 0 has equation y = x + 1.
Prove that eˣ ≥ x + 1 for every real x, by studying f(x) = eˣ - x - 1.
f′(x) = eˣ - 1
I differentiate: the derivative of eˣ is eˣ, that of -x is -1, and the constant vanishes.
eˣ - 1 = 0 ⟺ x = 0
Since exp is strictly increasing and equals 1 at 0: f′ is negative for x < 0 and positive for x > 0.
f(0) = e⁰ - 0 - 1 = 0
f decreases then increases: its minimum is f(0) = 0.
f(x) ≥ 0 donc eˣ ≥ x + 1
The minimum of f is 0, so f is nonnegative everywhere: the exp curve stays above its tangent y = x + 1.
Result: For every real x, eˣ ≥ x + 1, with equality only at x = 0.