I have always had difficulty when rearranging large equations of functions to find the roots and also the turning points(using derivatives) and was hoping some people could give me some tips when it comes to arranging algebra equations to solve them.
For example i am given the equation $$f(x)= {e^x \over (2x^2+1)}-1/2$$
We are told to find the y intercept, i know to set the left hand side of the equation to 0 but then i find it daunting to rearrange the right hand side, and am not sure on what order to start getting rid of elements, like does BEDMAS still apply or do i start buy adding 1/2 to each side then multiplying 1 to get rid of the 1 from the bottom.
And if i want to find the turning points of the function i know to find the derivative and then solve for y=0
$$f'(x) = {(2x^2-4x+1)e^x\over(2x^2+1)^2}$$
$$0 = {(2x^2-4x+1)e^x\over(2x^2+1)^2}$$
this is even scarier than the first when it comes to solving/rearranging
Any tips on this struggle of mine greatly appreciated
On your second example, it is far less scary than it looks. The first thing you generally want to do is try to get rid of the fractions by multiplying through by the denominators, and in this case, that gives you: $$(2x^2-4x+1)e^x = 0(2x^2+1)^2 = 0$$
Now, remember that the only way a product can be zero is if at least one of its factors is 0: $$ (2x^2 - 4x + 1) = 0 \qquad\text{or}\qquad e^x = 0$$ The latter equation can never be true, so the former one must be. which you can solve by the quadratic equation.
For solving $${e^x \over (2x^2+1)}-1/2 = 0$$ multiply through by the least common multiple of the two denominators $2(2x^2+1)$: $$2e^x - 1(2x^2 + 1) = 0$$ then move to the other side: $$2e^x = 2x^2 + 1$$ Here you should see that you are trying to equate two different kinds of functions, and realize that a nice neat solution just isn't possible. Thus JDThinking's comment that it's time to pull out Newton's method.