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.

**Read Online or Download Bob Miller's High School Calc for the Clueless - Honors and AP Calculus AB & BC PDF**

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**Additional info for Bob Miller's High School Calc for the Clueless - Honors and AP Calculus AB & BC **

**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.