Linear Approximation It is a way of finding the equation of a line tangent to a curve and using it to approximate a y value of the curve.

Method #1 Using f(x) = x³ + 4x – 1, approximate f(3.001). ~ f ‘(x) = 3x² + 4 1)Choose nice number (most likely whole)  a = 3 2)dx = given value – a  – 3 =.001 3)dy = f ‘(a) times dx  (3(3²) + 4) x (.001) = )f(3.001) = f(a) + dy  (3³ + 4(3) – 1) =

Method #2 Approximate f (2.1). f(x) = x³ – 2x + 3. ~ f ‘(x)= 3x² – 2 a)nice number  x = 2  y = (2³ - 2(2) + 3) = 7 slope: f ‘ (2) = 10 b)point slope: y-y 1 = m ( x- x 1 )  y – 7= 10 (x – 2) y = 10x – 13 c)approx. with given y-value  y = 10(2.1) – 13 y = 8

Newton’s Method Using a tangent line to find the zeros (x-intercepts, roots) of a given function. ~ Find the slope of the tangent line. Slope: f ‘(x n ) = f (x n ) x n - x n + 1 hint: x n + 1 is the next number Therefore: x n + 1 = x n - f (x n ) f ‘(x n )

Example: f(x) = x³ - x – 1; x 0 = 1.5, find x 2. ~ f ‘(x) = 3x² - 1 x 1 = x 0 – [f(x 0 ) / f ‘(x 0 )]  x 1 = 1.5 – ( / 5.75) = about x 2 = x 1 – [f(x 1 ) / f ‘ (x 1 )]  x 2 = – (0.101 / 1.030) = about Best way to calculate these numbers is to store them in your calculator. *graphing calc: press “STO  ” button, then select variable to store as.

Graphing Calculator Shortcut!! Example: f(x) = x³ + 4x –1, find root using Newton’s Method First, graph the function and recognize a whole number the zero is close to. Remember it! In this case, it is 0. \Y 1 = x^3+4x-1  unhighlight equal sign here by pressing “ENTER” on it. Scroll down to bottom of screen… \Y 0 = x - Y 1 / nDeriv( Y 1,x,x)  equal sign should be highlighted here

Go to “TBLSET” by pressing “2 nd” then “WINDOW” Select “Ask” for Indpnt. Go to “TABLE” by pressing 2 nd “GRAPH” The table should be blank Remember that number you chose before? For this equation, 0. Type that in for X and press “ENTER”. You get.25 for Y 0. Then enter.25 for the next X. You get Then enter for the next X. You get for Y 0 again. The zero for this problem is at where x= ~The zero is at where you get the same Y 0 on the table twice.

Reimann’s Sums They are used to approximate the area under a curve. Example on the right is y = x² [0,4]

Left - width of each rectangle times (sum of each height) 1 ( ) = 14

Right - width of each rectangle (sum of each height) 1 ( ) = 30

Midpoint- width of each rectangle (sum of each height) 1 [ (1/4) + (9/4) + (25/4) + (49/4)] = 21

Trapezoid area of trapezoid = (1/2)(width)(b 1 + b 2 ) use similar equation like below (1/2)(1)( ) = 22