solve for the second derivative and set to zero

Question

f"(x) = ((x^2-4)*sin(2x)- [(2x)(1+cos^2(x))]/ ((1+cos^2(x))^2)

i don't know how to set it to zero and solve i get this:

(x^2-4)(sin(2x))= (2x)(1+cos^2(x))

thanks for help

i did not get a clear response earlier so i am reposting. thanks

Answer 1

what i wrote is second derivative. i just need to set it to zero and solve, but that is where i am stuck

Answer 2

Hmmm. Parentheses for f"(x) are not balanced, but if your second equation is true,

(x^2-4)(sin(2x)) = (2x)(1+cos^2(x))

There's no easy analytical solution for this, but it's true for x=0. There are lots of other solutions, but they're harder to come up with.

Answer 3

I think you have trouble solving this because you didn't simplify second derivative correctly after you set it equal to zero.

Is (1+cos^2(x))^2 a denominator only for 2x or for the whole expression (x^2-4)*(sin(2x)-2x(1+cos^2(x)?

If it's a denominator for the whole expression then after I set it equal to zero I get [(x^2-4)*sin(2x)-2x]/[1+cos^2(x)] and since it's a fraction you can set numerator equal to zero (x^2-4)*sin(2x)-2x=0 and then set denominator equal to zero 1+cos^2(x)=0. And solve both equations for x.

Answer 4

Setting the denominator to zero is no good, because you cannot divide by zero.

If the numerator is zero, then the fraction is zero. (as long as the denominator is not also zero!)

0/10 = 0
10/0 is undefined

This puppy is just not open to algebraic solution.

Answer 5

Steeve you are right - I shouldn't have set denominator to zero but not because of your reasoning. No. But because it's a second derivative. When it comes to first derivative you CAN set (and have to sometimes) denominator to zero if you want to find vertical asymptotes!

But since here it is a second derivative setting numerator to zero is all that needs to be done.

Still, simplify the second derivative correctly and you'll have no problem solving it