Download Bob Miller's Calc for the Clueless: Calc I by Bob Miller PDF

By Bob Miller

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Additional info for Bob Miller's Calc for the Clueless: Calc I

Example text

Solving for ∆x, ∆x = (6 - 3)/n = 3/n. 2. From before, the approximate sum is f(w1)∆x1 + f(w2)∆x2 + f(w3)∆x3 + ... + f(wn)∆xn. All of the ∆xi are equal to ∆x, and we will take the right end of each interval as the point where we will take the height. , and wn = xn. 3. Rewriting 2, we factor out the ∆x and get [f(x1) + f(x2) + f(x3) + ... + f(xn)∆x. 4. Now f(x) = x2 + 4x + 7. 5. Our task is now to add all these up and then multiply everything by ∆x. A. If we multiply out f(x1), we see that the number we get from this term is 32 + 4(3) + 7 = 28.

Example 15— Vertical asymptotes at x = 4, f(4-), f(4+) are positive, and the curve near x = 4 looks like... 00000001 = 700,000,000, which is big. The curve tends to plus infinity, from the right side of 4. Similarly, the curve goes to plus infinity from the left side. Example 16— Vertical asymptote at x = 3. f(3 -) and f(3+) are negative. The curve near x = 3 looks like... To summarize, if the exponent is even positive in the denominator, near the vertical asymptote, both ends go to either plus infinity or minus infinity.

1. f'(2)= infinity 2. f(2) = 3 (2,3) 3. f'(2-) is negative, f'(2+) is positive. Cusp with the point down. -3 = (x - 2)4/5. (x - 2)1/5 = ±(-3)1/4, which is imaginary. No x intercepts, y intercept: x = 0. y = (-2)4/5 + 3. [0,(-2)4/5 + 3] = A. No round max or min and no inflection points, f(1000) is positive and f(-1000) is positive. Both ends go to plus infinity. The sketch: Example 28— Intercept (0,0). No round max or min and no inflection points. 1. |f'(0)| = infinity 2. f(0) = 0 3. f'(0-), f'(0+) both positive.

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