Bob Miller's High School Calc for the Clueless - Honors and by By (author) Bob Miller

By By (author) Bob Miller

With Bob Miller at your aspect, you by no means must be clueless approximately math back!

Algebra and calculus are tricky on highschool scholars such as you. Professor Bob Miller, with greater than 30 years' educating adventure, is a grasp at making the complicated basic, and his now-classic sequence of Clueless learn aids has helped tens of millions comprehend the cruel subjects.

Calculus-with its integrals and derivatives-is recognized for tripping up even the fastest minds. Now Bob Miller-with his 30-plus years' event educating it-presents highschool calculus in a transparent, funny, and fascinating way.

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Example text

Since the limit of f(x) is L, at some point, L ϩ 1 must be bigger than f(x). 2. Since lim gsxd ϭ M, given an ε/2(|L| ϩ 1), there xSa exists a ␦2 such that whenever 0 Ͻ |x Ϫ a| Ͻ ␦2, then |g(x) Ϫ M| Ͻ ε/2(|L| ϩ 1). 3. Again since lim fsxd ϭ L, given an ε/2|M|, there xSa exists a ␦3 such that whenever 0 Ͻ |x Ϫ a| Ͻ ␦3, then |f(x) Ϫ L| Ͻ ε/2|M|. Now let ␦ ϭ minimum (␦1, ␦2, ␦3). So |f(x)| |g(x) Ϫ M| ϩ |M| ϩ |f(x) Ϫ L| Ͻ (|L| ϩ 1) ε/2(|L| ϩ 1) ϩ |M|ε/2|M| ϭ ε/2 ϩ ε/2 ϭ ε. Whew! As I said, the techniques will come sooner; the understanding will come later.

All x values on P2Q are the same. The length of P2Q ϭ f(x ϩ ⌬x) Ϫ f(x). The slope of the secant line is L1 ϭ P2Q ⌬y f(x ϩ ⌬x) Ϫ f(x) ϭ ϭ P1Q ⌬x ⌬x Now we do as before—let P2 go to P1. Algebraically this means to take the limit as ⌬x goes to 0. We get the slope of the tangent line L2 at P1. Our notation will be the slope of the tangent line L2 ϭ lim ⌬x S 0 f(x ϩ ⌬x) Ϫ f(x) ⌬x if it exists. DEFINITION Suppose y ϭ f(x). The derivative of f(x), at a point x denoted b, fЈ(x), or dy/dx, is defined as f(x ϩ ⌬x) Ϫ f(x) ⌬x S 0 ⌬x lim if it exists.

If the tops are the same, the larger the bottom, the smaller the fraction. 3/10 Ͼ 3/11. Now let’s do some problems. EXAMPLE 18— Using ε and ␦, prove that lim (4x Ϫ 3) ϭ 5 xS2 In the definition lim f(x) ϭ L xSa f(x) ϭ 4x Ϫ 3, a ϭ 2, and L ϭ 5. Given ε Ͼ 0, we must find ␦ Ͼ 0, such that if 0 Ͻ |x Ϫ 2| Ͻ ␦, then |(4x Ϫ 3) Ϫ 5| Ͻ ε. |(4x Ϫ 3) Ϫ 5|ϭ|4x Ϫ 8|ϭ|4(x Ϫ 2)| ϭ|4||x Ϫ 2| Ͻ 4 . ␦ ϭ ε ␦ ϭ ε/4 The Beginning—Limits EXAMPLE 19— Prove lim (x2 ϩ 2x) ϭ 24 xS4 Given ε Ͼ 0, we must find ␦ Ͼ 0, such that if 0 Ͻ |x Ϫ 4| Ͻ ␦, then |x2 ϩ 2x Ϫ 24| Ͻ ε.

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