Mindrust

So for the third piece of g(x), I was missing the constant at the end, C. To find C, I let the third equal the second, at x=1. So... 2x - 0.5x^2 + C = 0.5x^2 (at x=1) Then solve for C, which is -1. So the answer to the third is 2x - 0.5x^2 - 1 I know I have the right answer, but not sure if my method is correct. Can anyone confirm? Thanks!

Not sure what I'm doing wrong, but I can't seem to get the integral for the third and fourth piece. I might be using the wrong values for [a,b]. Anyone know the answers and what intervals should be used? Am I supposed to add the first two parts into the third?

Thanks. BTW, what does that big "pi" looking symbol mean?

Find the nth derivative of sqr(2x-1). That means we need to find a pattern. Here's what I've got so far, from 1st derivative to 5th: The pattern for the exponent is simple, 1/2 - n The pattern for the co-efficient is what I'm stuck on. Here's what it looks like. I'm not sure how to express it mathematically. 1 -> 1 = 1 2 -> -1 = 1(-1) 3 -> 3 = 1(-1)(-3) 4 -> -15 = 1(-1)(-3)(-5) 5 -> 105 = 1(-1)(-3)(-5)(-7) and so on... Anyone know how to make a function for the coefficient in terms of n?

You're right, what I meant to say was: as x approaches zero, 1/x = infinity.

How does one go about solving this algebraically? I tried creating a new variable (let a = log(base5)x) to get a^(1/2) + a^(1/3) = 2 but not sure what to do from there... If I can figure out how to get rid of the radicals, the rest should be simple. [EDIT ] A friend helped me with the answer. Let y = log_5(x)Let z = y^(1/6) Then the given equation becomes z^3 + z^2 = 2, that is (z-1)(z+1+i)(z+1-i) = 0. The solutions are r = 1, s = -1-i, and t=-1+i. Then x = 5^(r^6) =, 5^(s^6), 5^(t^6) are solutions of the original equation. The only real valued solution is x = 5. Alternatively, you can arrange the equation into cubic form, and get the roots using the cubic formula.

The concept looked interesting at first, but several problems emerged after thinking about it. First, 1/0 = undefined ; 1/(number close to zero) = infinity That means there must be at least something, an infinitely small amount of energy, if you will. The second problem is, even if we say 1/0 = infinity, then (1/0) x 0 = infinity x 0 = 1 x 0 = infinity x 0 = infinity x 0 = 0 Why? Because 0/0 does not equal to 1 as you assume in your initial calculation. 0/0 = 0. One could even say it's undefined. Third, if you want to get really technical, 0 does not equal nothing, in mathematics. There is another symbol to represent that.