Exponential Functions

Definition: An exponential function is of the form f(x) = Cax with the base a > 0 and a 6= 1, C is a real number.

Recall: • a0 = 1 • a1 = a • a2 = a · a, ...

−n 1 • a = an

1 √ th n • a n = n root of a = a

p p( 1 ) 1 1 • a q = a q = (ap) q = (a q )p

Rules of Exponents: • anam = an+m Example: 52 · 53 = (5 · 5)(5 · 5 · 5) = 55 • (an)m = anm Example: (52)3 = (52)(52)(52) = 56

an n−m 53 −4 • am = a Example: 57 = 5 • (ab)n = anbn Example: 65 = (2 · 3)5 = 25 · 35

a n an 2 5 25 • ( b ) = bn Example: ( 3 ) = 35

In the following problems, simplify to an integer or a single exponent an: √ 1 √ 3 √ 1 3 √ 4 √ 2 2 2 + 2 2 2 1 2 2 1 • ( 2 )( 2 ) = 2 = 2 = 2 = (2 2 ) = 2 2 = 2 = 2 √ √ √ √ • (2 3) 3 = 2 3· 3 = 23 = 8 √ √ √ √ • (32+ 5)(32− 5) = 32+ 5+2− 5 = 34 = (32)2 = 92 = 81

Property: If a 6= 1, ax = ay ⇐⇒ x = y. Therefore the function bx is a 1-1 function. Solve for x: • 27x = 9 ⇒ (9 · 3)3 = 9 ⇒ (33)x = 32 3x 2 2 ⇒ 3 = 3 ⇐⇒ 3x = 2 ⇐⇒ x = 3

• 8x+1 = √32 2 x+1 − 1 3 2 − 1 3+2− 1 9 ⇒ 8 = (8 · 4)(2 2 ) = (2 · 2 )(2 2 ) = 2 2 = 2 2 x+1 9 ⇒ 8 = 2 2 3 x+1 9 ⇒ (2 ) = 2 2 3(x+1) 9 ⇒ 2 = 2 2 3x+3 9 ⇒ 2 = 2 2 9 ⇐⇒ 3x + 3 = 2 ⇒ 2(3x + 3) = 9 ⇒ 6x + 6 = 9 ⇒ 6x = 3 3 1 ⇒ x = 6 ⇒ x = 2 1 Review of Theorems and Formulas for Exam 2 All exams will consist solely of homework type problems. For odd degree vertical asymptotes, one side goes to +5, the other to All numeric answers must be exact, no decimals, no mixed -5. 3 1 fractions. E.g., 2 ,2 not, 1 2 or 1.5 or 1.414... . For even degree vertical asymptotes, both sides go to +5 or both go DEFINITION. f -1, the inverse of f , is the function, if any, such that, to -5. f (f -1(x)) = x when f -1(x) is defined and For a reduced rational function: f -1(f (x)) = x when f (x) is defined. } x-intercepts (roots) occur where the top is 0. THEOREM. The graph of y = f -1(x) is the reflection of the graph of y = If the root has degree n, the x-intercept looks like that of y = xn or y f (x) across the major diagonal y = x. = -xn . DEFINITION. f is 1-1 (“one-to-one”) iff } If the bottom is 0 at a, then x=a is a vertical asymptote. x = y implies f (x) = f (y). If the factor has degree n, the vertical asymptote looks like that of THEOREM. f has an inverse iff f is 1-1 iff no horizontal line y =1/xn or y = -1/xn . intersects its graph more than once. } As xƒ+5, the graph resembles the graph of the leading term DEFINITION. A quadratic function is a degree-2 polynomial which is either a constant b or of the form a/xn or axn. 2 y ax bx c with a=0. (1) If a constant b, then y = b is a horizontal asymptote. The graph is a parabola. (2) If it is a/xn, then y = 0 is a horizontal asymptote. } If a>0, the horns point up. (3) If it is axn, there is no horizontal asymptote. } If a<0, the horns point down. DEFINITION. An exponential function is of the form y = bx with the } If |a|>1, the parabola is narrower than y = x2. base b>0. } If |a|<1, the parabola is wider than y = x2. b0 = 1, b1 = b, b2 = b.b, ... b-n = 1/bn 1/n n th p/q p.(1/q) p 1/q 1/q p THEOREM. Every quadratic function may be written in the form: b = b , the n root of b b = b = (b ) = (b ) 2 y ax x0 y0 EXPONENT RULES Examples n m nm 2 3 6 where (x0, y0) is the vertex (nose) of the parabola. b b 5 5 For an expanded polynomial axn ... c with axn the term of highest bnbm bnm 5254 56 7 degree: axn is the leading term, a is the leading coefficient, and n bn nm 53 4 1 5 4 ***Inbm theb following example57 5 the variable54 , 53 is in5 the base rather than the exponent. is the degree. The y-intercept is the constant term c. n n n8 9 5 5 5 Example:ab a bz = 2 2 2 3 6 y=x y=x2 2 n 5 y=-x a n8 1 a 9 1 2 5 2 ⇒ b(z )8 b=n (2 2 ) 8 3 35 9 x y x P⇒ROPERTYz = 2. 16If b = 1, b b x y . Hence b is 1-1. a b means a is approximately equal to b. 10 3 10 3 Fact: 2 10 , i.e., 2 x is approximately equal to 10 . x The graph of y =x 1 is the horizontal line y = 1. Otherwise, the graph of y = a 4 The graph of y = 1 is the horizontal line y = 1. 3 y=x 3 y=x y=-x Otherwise,1. has the y-intercept graph of y 1 = but bx NO x-intercept, } has y-intercept 1 but no x-intercept, } it2. goesgoes to to5 ∞in onein one direction, direction, it has the horizontal asymptote y = 0 in the other. n } 3. has the horizontal asymptote y = 0 in the other direction. } For large x (near +5), graph looks like the leading term ax . For b >1, the graph of bx is like the graph of 2x as below. } As x goes to 5, y goes to +5 if a>0, to -5 if a<0. x x 1 x −x ForExample: 0 1, the graph crosses the x-axis like y=x3 1 or y=-x3 y=0 hor asymp } At roots of even degree, the graph touches but doesn't cross the x x x-axis, like y=x2 or y=-x2 DEFINITION. e is the unique exponential function b whose the tangent at (0,1) has slope 1. Fact: e 2.7 DEFINTION. A ratio of two polynomial functions is a rational function. We can use techniques of translations and reflections we’ve already seen to graph exponential DEFINITION. log x , the log of x to the base b, is the inverse of the It is reduced if the top and bottom have no common factors. functions: b exponential function bx. `In the graphs below, x = 0 (the y-axis) is the vertical asymptote, y = ln(•x),Graph the naturalf(x) logarithm = 2−x , = log x =theReflect inverse2 ofx acrossex. y-axis. 0 (the x-axis) is the horizontal asymptote. e Inverses act in opposite directions and inverses cancel. t 2 x−1 y=1/x y=1/x 3 • Graph f(x) =y 2 − 1 Shift right 1,y down 2. y=1/x y logbx iff b xe . y = ln(x) iff x . x log x x ln(x) x=0 log b x. b b x . ln(e ) = x. e =x. vert b y=0 x y=0 y=0 If the exponent base is b instead of e, ln and b don’t completely x hor x=0 x=0 Thecancel: function the bottome isleft the equation unique becomes: exponential function whose tangent at (0,1) has slope 1. lnbxx ln b. The exponent comes down to the outside. 3 Fact: e ≈ 2.7 y= -1/x y= -1/x2 y=-1/x FACT. e0 = 1ˆ 0 = ln(1). e1 = e ˆ 1 = ln(e) x=0 y=0 `Know area, circumference, volume formulasx for triangles, x x y=0 Hence 2 < e < 3 ⇒ the graph of e lies between the graphs of 2 and 3 . hor y=0 rectangles, boxes and cans. See inside front cover of the text. x=0 vert x=0 x 2 −x2 (e ) Example: Solve for x: e = e3 .

x 2 2 (e ) e−x = e3 e2x = e3 = e2x · e−3 = e2x−3

2 So e−x = e2x−3 ⇒ −x2 = 2x − 3 ⇒ x2 + 2x − 3 = 0

⇒ (x + 3)(x − 1) = 0 ⇒ x = −3, 1.

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