# What are the points of inflection of #f(x)=8x^2sin(2x-pi) # on # x in [0, 2pi]#?

By signing up, you agree to our Terms of Service and Privacy Policy

To find the points of inflection, first find the second derivative of the function ( f(x) = 8x^2 \sin(2x - \pi) ). Then, solve for ( x ) when the second derivative equals zero. Afterward, check the concavity of the function around those points to confirm if they are points of inflection.

The second derivative of ( f(x) ) is given by:

[ f''(x) = 32\sin(2x - \pi) - 64x\cos(2x - \pi) ]

To find points of inflection, set ( f''(x) = 0 ) and solve for ( x ). Then, determine the concavity of ( f(x) ) around those points.

[ 32\sin(2x - \pi) - 64x\cos(2x - \pi) = 0 ]

Solve for ( x ) within the interval ( [0, 2\pi] ). After finding these points, check the concavity of ( f(x) ) around them to confirm if they are points of inflection.

By signing up, you agree to our Terms of Service and Privacy Policy

- If #f(x)=sec(x)#, how do I find #f''(π/4)#?
- What are the points of inflection, if any, of #f(x)= -14x^3 + 19x^2 - x - 2 #?
- How do you find local maximum value of f using the first and second derivative tests: #f(x) = x + sqrt(9 − x) #?
- If #f(x)=x(x^2+1) #, what are the points of inflection, concavity and critical points?
- What are the points of inflection, if any, of #f(x)=1/x #?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7