Intuition The big pictureAn exponential function is repeated multiplication turned continuous . Instead of asking "add the same amount each step" (that's a line ), we ask "multiply by the same factor each step". That single change — multiply not add — is what makes a x a^x a x
Definition Exponential functionA function of the form
f ( x ) = a x , a > 0 , a ≠ 1 f(x) = a^x, \qquad a > 0,\ a \neq 1 f ( x ) = a x , a > 0 , a = 1 a a a base (a fixed positive constant) and x x x x x x y > 0 y > 0 y > 0
WHY the restrictions?
a > 0 a > 0 a > 0 a < 0 a < 0 a < 0 a 1 / 2 = a a^{1/2} = \sqrt{a} a 1/2 = a a ≠ 1 a \neq 1 a = 1 1 x = 1 1^x = 1 1 x = 1 x x x We allow a = a= a = 2 x 2^x 2 x 10 x 10^x 1 0 x ( 1 / 3 ) x (1/3)^x ( 1/3 ) x e x e^x e x
Start from what we already know and extend by demanding one law stays true : the index law
a m + n = a m ⋅ a n . a^{m+n} = a^m \cdot a^n. a m + n = a m ⋅ a n .
Positive integers: a 3 = a ⋅ a ⋅ a a^3 = a\cdot a\cdot a a 3 = a ⋅ a ⋅ a Zero: we need a 0 ⋅ a n = a 0 + n = a n a^0 \cdot a^n = a^{0+n}=a^n a 0 ⋅ a n = a 0 + n = a n a 0 = 1 a^0 = 1 a 0 = 1 Why? To keep the law consistent.Negatives: need a − n ⋅ a n = a 0 = 1 a^{-n}\cdot a^n = a^0 = 1 a − n ⋅ a n = a 0 = 1 a − n = 1 a n a^{-n} = \dfrac{1}{a^n} a − n = a n 1 Fractions: need ( a 1 / 2 ) 2 = a 1 = a (a^{1/2})^2 = a^{1} = a ( a 1/2 ) 2 = a 1 = a a 1 / 2 = a a^{1/2}=\sqrt a a 1/2 = a a p / q = a p q a^{p/q}=\sqrt[q]{a^p} a p / q = q a p Irrationals (like a 2 a^{\sqrt2} a 2 2 \sqrt2 2 1.41 , 1.414 , … 1.41,1.414,\dots 1.41 , 1.414 , …
Intuition Why this mattersEvery property of the graph flows from one demand : the multiplication law must hold for all real exponents. The graph is just the visual shadow of that algebra.
Take a > 1 a>1 a > 1 x → − ∞ x\to -\infty x → − ∞ x = − N x=-N x = − N N → + ∞ N\to+\infty N → + ∞ a − N = 1 a N . a^{-N} = \frac{1}{a^N}. a − N = a N 1 . a > 1 a>1 a > 1 a N → ∞ a^N \to \infty a N → ∞ 1 a N → 0 + \dfrac{1}{a^N}\to 0^+ a N 1 → 0 + arbitrarily close to 0 0 0 asymptote . It never equals 0 0 0 1 / ( finite positive ) 1/(\text{finite positive}) 1/ ( finite positive ) 0 0 0
For 0 < a < 1 0<a<1 0 < a < 1 x → + ∞ x\to+\infty x → + ∞
Intuition Growth vs decay = a single mirror( 1 / a ) x = a − x (1/a)^x = a^{-x} ( 1/ a ) x = a − x y = ( 1 / a ) x y=(1/a)^x y = ( 1/ a ) x y = a x y=a^x y = a x reflected in the y y y . Decay is just growth run backwards.
Worked example 1 — Sketchy = 2 x y=2^x y = 2 x Points: x = 0 ⇒ 1 x=0\Rightarrow 1 x = 0 ⇒ 1 x = 1 ⇒ 2 x=1\Rightarrow2 x = 1 ⇒ 2 x = 2 ⇒ 4 x=2\Rightarrow4 x = 2 ⇒ 4 x = − 1 ⇒ 1 2 x=-1\Rightarrow \tfrac12 x = − 1 ⇒ 2 1 x = − 3 ⇒ 1 8 x=-3\Rightarrow\tfrac18 x = − 3 ⇒ 8 1 Why these steps? We pick the two anchors ( 0 , 1 ) , ( 1 , 2 ) (0,1),(1,2) ( 0 , 1 ) , ( 1 , 2 ) see the curve flattening toward 0 0 0 x → − ∞ x\to-\infty x → − ∞ 2 x → 0 + 2^x\to0^+ 2 x → 0 + y = 0 y=0 y = 0
Worked example 2 — Comparey = 2 x y=2^x y = 2 x y = 5 x y=5^x y = 5 x ( 0 , 1 ) (0,1) ( 0 , 1 ) x = 1 x=1 x = 1 2 2 2 5 5 5 Why this step? The base is the height at x = 1 x=1 x = 1 5 x 5^x 5 x cross only at ( 0 , 1 ) (0,1) ( 0 , 1 ) .
Worked example 3 — Decay:y = ( 1 3 ) x y=\left(\tfrac13\right)^x y = ( 3 1 ) x ( 1 3 ) x = 3 − x \left(\tfrac13\right)^x = 3^{-x} ( 3 1 ) x = 3 − x Why? This instantly tells us it's 3 x 3^x 3 x y y y x = 0 ⇒ 1 x=0\Rightarrow1 x = 0 ⇒ 1 x = 1 ⇒ 1 3 x=1\Rightarrow\tfrac13 x = 1 ⇒ 3 1 x = − 1 ⇒ 3 x=-1\Rightarrow3 x = − 1 ⇒ 3 y > 0 y>0 y > 0 y = 0 y=0 y = 0
Worked example 4 — A transformation:y = 2 x + 3 y=2^x+3 y = 2 x + 3 up 3 . Why? Adding 3 to every output raises each point. New asymptote: y = 3 y=3 y = 3 0 0 0 y y y = 2 0 + 3 = 4 =2^0+3=4 = 2 0 + 3 = 4 y > 3 y>3 y > 3
a x a^x a x Why it feels right: on the graph the curve looks like it touches the x x x y = 0 y=0 y = 0 The fix: it only approaches 0 0 0 a x = 1 / a N a^x=1/a^N a x = 1/ a N 0 0 0 y = 0 y=0 y = 0 limit, not a value . Range is strictly y > 0 y>0 y > 0
a x a^x a x x a x^a x a Why it feels right: both have a power. But x 2 x^2 x 2 parabola ; 2 x 2^x 2 x exponential . The exponent is what varies in an exponential.
Common mistake Thinking bigger base ⇒ highery y y Why it feels right: bigger numbers seem "higher." Fix: ALL a x a^x a x ( 0 , 1 ) (0,1) ( 0 , 1 ) a 0 = 1 a^0=1 a 0 = 1 steepness , not the intercept.
y = 2 x + 3 y=2^x+3 y = 2 x + 3 y = 0 y=0 y = 0 Fix: vertical shifts move the asymptote too. The floor rose to y = 3 y=3 y = 3
Recall Try before reading answers
What point does every y = a x y=a^x y = a x
Why is the range y > 0 y>0 y > 0
Derive the asymptote of 2 x 2^x 2 x x → − ∞ x\to-\infty x → − ∞
How does y = ( 1 / a ) x y=(1/a)^x y = ( 1/ a ) x y = a x y=a^x y = a x
Recall Feynman: explain to a 12-year-oldImagine a magic bacteria that doubles every hour . Start with 1. After 1 hour: 2, then 4, 8, 16… it shoots up crazy fast — that's 2 x 2^x 2 x backwards in time: one hour ago there was half, before that a quarter, an eighth… it gets tinier and tinier but never actually hits zero (you always have some bacteria, even a speck). That "never quite reaching zero" floor is the asymptote . And no matter what number you double or triple, at "hour zero" you always start with 1 — that's why every curve goes through ( 0 , 1 ) (0,1) ( 0 , 1 )
"POWER ON TOP → SHOOTS OFF THE TOP." The variable is on top (the exponent), so the graph takes off — and its feet never leave the floor y = 0 y=0 y = 0
What is the general form of an exponential function and the base restrictions? f ( x ) = a x f(x)=a^x f ( x ) = a x with
a > 0 , a ≠ 1 a>0,\ a\neq1 a > 0 , a = 1 .
Why must a > 0 a>0 a > 0 a x a^x a x Negative bases give non-real values like
a 1 / 2 = a a^{1/2}=\sqrt a a 1/2 = a ; positivity keeps it real for all
x x x .
Why is a ≠ 1 a\neq1 a = 1 1 x = 1 1^x=1 1 x = 1 is a flat line, not a genuine exponential.
What point do all y = a x y=a^x y = a x ( 0 , 1 ) (0,1) ( 0 , 1 ) , because
a 0 = 1 a^0=1 a 0 = 1 for every base.
What is the coordinate that reveals the base directly? ( 1 , a ) (1,a) ( 1 , a ) , since
a 1 = a a^1=a a 1 = a .
What is the range of a x a^x a x y > 0 y>0 y > 0 (strictly positive, never zero).
What is the horizontal asymptote of y = a x y=a^x y = a x y = 0 y=0 y = 0 (the
x x x -axis).
Prove 2 x → 0 2^x\to0 2 x → 0 x → − ∞ x\to-\infty x → − ∞ 2 − N = 1 / 2 N 2^{-N}=1/2^N 2 − N = 1/ 2 N ; as
N → ∞ N\to\infty N → ∞ ,
2 N → ∞ 2^N\to\infty 2 N → ∞ , so
1 / 2 N → 0 + 1/2^N\to0^+ 1/ 2 N → 0 + .
Why does the curve never touch y = 0 y=0 y = 0 a x = 1 / a N a^x=1/a^N a x = 1/ a N is a positive fraction; a nonzero-denominator fraction is never 0.
How is y = ( 1 / a ) x y=(1/a)^x y = ( 1/ a ) x y = a x y=a^x y = a x ( 1 / a ) x = a − x (1/a)^x=a^{-x} ( 1/ a ) x = a − x , a reflection of
a x a^x a x in the
y y y -axis (decay = reversed growth).
Difference between a x a^x a x x a x^a x a a x a^x a x is exponential (variable in exponent);
x a x^a x a is a power/polynomial (variable in base).
What happens to the asymptote of y = 2 x + 3 y=2^x+3 y = 2 x + 3 It shifts up to
y = 3 y=3 y = 3 ; range becomes
y > 3 y>3 y > 3 ; intercept
( 0 , 4 ) (0,4) ( 0 , 4 ) .
Growth vs decay condition? a > 1 a>1 a > 1 ⇒ increasing (growth);
0 < a < 1 0<a<1 0 < a < 1 ⇒ decreasing (decay).
Why is a 0 = 1 a^0=1 a 0 = 1 From
a 0 ⋅ a n = a 0 + n = a n a^0\cdot a^n=a^{0+n}=a^n a 0 ⋅ a n = a 0 + n = a n , dividing gives
a 0 = 1 a^0=1 a 0 = 1 .
Logarithms as the inverse of exponentials — reflect a x a^x a x y = x y=x y = x log a x \log_a x log a x The number e and natural exponential eˣ — the special base where slope = height.Index laws — the algebra that defines a x a^x a x x x x Exponential growth and decay models — real-world use (population, radioactivity).Graph transformations — shifts/reflections applied to a x a^x a x
else non-real or flat line
Intuition Hinglish mein samjhoDekho, exponential function a x a^x a x multiply karte hain. Isi ek chhoti si difference se graph tezi se upar chala jaata hai (agar a > 1 a>1 a > 1 0 < a < 1 0<a<1 0 < a < 1 a a a 1 1 1 a \sqrt a a 1 x 1^x 1 x
Sabse important baat: har a x a^x a x ( 0 , 1 ) (0,1) ( 0 , 1 ) a 0 = 1 a^0=1 a 0 = 1 ( 1 , a ) (1,a) ( 1 , a ) y > 0 y>0 y > 0 x x x asymptote y = 0 y=0 y = 0
Asymptote ka logic simple hai: 2 − N = 1 / 2 N 2^{-N} = 1/2^N 2 − N = 1/ 2 N N N N 2 N 2^N 2 N 1 / 2 N 1/2^N 1/ 2 N ( 1 / a ) x = a − x (1/a)^x = a^{-x} ( 1/ a ) x = a − x y y y
Exam tip: a x a^x a x x a x^a x a x 2 x^2 x 2 2 x 2^x 2 x + 3 +3 + 3 y = 3 y=3 y = 3
