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Since we are dealing with complex numbers and taking its modulus, we can rewrite (z-i)=((-1)(i-z))=(i-z) doing the same for (z-1)=(1-z) we get (i-z)+(z+i)+(1-z)+(z-1)=(i+i+z-z+1+1+z-z) =(2i+2)=4 as we are...

f'(x)=-2(3x+4)^(-3) * 3 = -6(3x+4)^(-3);
f''(x)= 18(3x+4)^(-4) * 3 = 54(3x+4)^(-4);
both found by using the chain rule for differentiation.

Then Maclaurin series up to x^2 is: f(x)=f(0)...

First, recognise that 3^2 = 9. Recall the rule for multiplying indices, that (a^b)^c = a^{bc}. Then, substitute 3^2 in place of 9 to get 3^{4x} = (3^2)^{(x-1)/2}. Use the rule for multiplying indices, so ...

Finding f(x) requires integrating the function f'(x), because f(x) is the integral of the given function f'(x). So {integralsymbol} f'(x) dx = {integralsymbol} (3x^2 - 5cos(3x) + 90) dx = x^3 - (5/3)sin(3...

This thing we have to solve is an inequality and the solution we are looking for is an entire range of real number, something like "every x between 1 and 2", for example. To do this we need to b...