Trig Derivatives (Curve Sketching)

[pahuja]

I have to graph a sketch of y=cos(x²) from -2π≤x≤2π using curve sketching methods. So far i have found x/y intercepts and critical points, but am stuck on the last step of finding the points of inflection that fall within the -2π≤x≤2π range.

I have calculated the second derivative and set it to 0, but don't know how to solve for "x":
y"= -2(sin(x²)+2x²cos(x²))
0 = -2(sin(x²)+2x²cos(x²))

 

[Deleted member 4993]

I have to graph a sketch of y=cos(x²) from -2π≤x≤2π using curve sketching methods. So far i have found x/y intercepts and critical points, but am stuck on the last step of finding the points of inflection that fall within the -2π≤x≤2π range.

I have calculated the second derivative and set it to 0, but don't know how to solve for "x":
y"= -2(sin(x²)+2x²cos(x²))
0 = -2(sin(x²)+2x²cos(x²))

Can you plot:

\(\displaystyle \theta \ = tan(\theta) \ \)

 

[Steven G]

I have to graph a sketch of y=cos(x²) from -2π≤x≤2π using curve sketching methods. So far i have found x/y intercepts and critical points, but am stuck on the last step of finding the points of inflection that fall within the -2π≤x≤2π range.

I have calculated the second derivative and set it to 0, but don't know how to solve for "x":
y"= -2(sin(x²)+2x²cos(x²))
0 = -2(sin(x²)+2x²cos(x²))

I am sure that you can do something to 0 = -2(sin(x²)+2x²cos(x²)). Like 0 = sin(x²)+2x²cos(x²), then sin(x²)=-2x²cos(x²)), then tan(x²)=-2x². Now what?

 

[Deleted member 4993]

I am sure that you can do something to 0 = -2(sin(x²)+2x²cos(x²)). Like 0 = sin(x²)+2x²cos(x²), then sin(x²)=-2x²cos(x²)), then tan(x²)=-2x². Now what?

tan(x²)=-2x² for cos(x²)\(\displaystyle \ \ne \ 0 \ \to \ x^2 \ \ne \frac{(2n+1)*\pi}{2}\)