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# Bob Miller's high school calculus for the clueless : high by Bob Miller

By Bob Miller

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

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

Calculus-with its integrals and derivatives-is well-known for tripping up even the fastest minds. Now Bob Miller-with his 30-plus years' adventure instructing it-presents highschool calculus in a transparent, funny, and interesting way.

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Additional info for Bob Miller's high school calculus for the clueless : high school calculus : honors calculus, AB and BC calculus

Example text

M ϭ yЈ ϭ 2x ϩ 4 ϭ 2(2) ϩ 4 ϭ 8. y ϭ (2)2 ϩ 4(2) ϩ 7 y Ϫ 19 ϭ 19. The line is 8 ϭ . xϪ2 C. m ϭ 2x ϩ 4. Since x2 ϩ 4x ϩ 7 ϭ 12, x2 ϩ 4x ϭ 5 ϭ 0 or (x ϩ 5)(x Ϫ 1) ϭ 0. So x ϭ 1 and x ϭ Ϫ5. The points are (1, 12) and (Ϫ5, 12). For the point (1, 12), m ϭ 2(1) ϩ 4 ϭ 6 and the y Ϫ 12 equation of the line is 6 ϭ x Ϫ 1 . For the point (Ϫ5, 12), m ϭ 2(Ϫ5) ϩ 6 ϭ Ϫ4 and y Ϫ 12 the equation is Ϫ4 ϭ x ϩ 5 . Here’s the picture: See, there are two lines where y ϭ 12. y (– 5, 12) L1 (1, 12) y = 12 L2 x The Basics D.

The example is then virtually the same as Example 15. Now seems to be a fine time to insert problems you are likely to see. They are written in roughly increasing order of difficulty. EXAMPLE 21— A. Find the equation of the line tangent to y ϭ x2 ϩ 4x ϩ 7 at the point (1, 12). B. Find the equation of the line tangent to y ϭ x2 ϩ 4x ϩ 7 if x ϭ 2. C. Find the equation of the lines tangent to y ϭ x2 ϩ 4x ϩ 7 when y ϭ 12. D. Find the equation of the line tangent to y ϭ x2 ϩ 4x ϩ 7 that is perpendicular to the line 5x ϩ 10y ϭ 11.

Since everything on P2Q has the same x value, the length of P2Q ϭ y2 Ϫ y1. The slope is mϭ y2 Ϫ y1 change in y ⌬y ϭx Ϫx ϭ ⌬x 2 1 change in x ⌬ ϭ delta, another Greek letter Let’s do the same thing for a general function y ϭ f(x). The Basics Let point P1 be the point (x, y) ϭ (x, f(x)). A little bit away from x is x ϩ ⌬x. ) The corresponding y value is f(x ϩ ⌬x). So P2 ϭ (x ϩ ⌬x, f(x ϩ ⌬x)). As before, draw a line through P1 parallel to the x axis and a line through P2 parallel to the y axis. The lines again meet at Q.