solve for the second derivative and set to zero
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
what i wrote is second derivative. i just need to set it to zero and solve, but that is where i am stuck
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.
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.
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.
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
To solve for the second derivative and set it to zero, we have the equation:
f''(x) = ((x^2-4) * sin(2x) - (2x)(1+cos^2(x))) / (1+cos^2(x))^2
Setting this equation to zero means finding the values of x that make f''(x) equal to zero.
To solve this equation, we will first simplify it by expanding and collecting like terms. Let's break it down step by step:
1. Expand the numerator:
((x^2-4) * sin(2x)) - (2x)(1+cos^2(x))
2. Distribute sin(2x) to both terms in the first parentheses for expansion:
(x^2-4) * sin(2x) = (x^2-4) * sin(2x)
3. Distribute -2x in the second parentheses for expansion:
(2x)(1+cos^2(x)) = 2x + 2x * cos^2(x)
Now, the equation becomes:
((x^2-4) * sin(2x)) - 2x - 2x * cos^2(x) = 0
To solve for x, you can use numerical methods or approximation techniques, as this equation does not have an analytical solution. One common approach is to use the Newton-Raphson method or a graphing calculator to find the roots of the equation.
Alternatively, you can plot the given function f''(x) = ((x^2-4) * sin(2x) - (2x)(1+cos^2(x))) / (1+cos^2(x))^2 and look for the values of x where the graph crosses the x-axis (y=0). This will give you approximate solutions to the equation.
Remember that setting the second derivative equal to zero helps us identify potential points of inflection in the original function f(x), where the concavity changes.