2nd Derivative Problem

[Jason76]

Post Edited

\(\displaystyle \dfrac{d^{2}y}{dx} = \sin(2x) - \cos(4x)\)

How many points of inflection does the graph of y(x) have on the interval of \(\displaystyle [0,10]\) ? Answer: seven

Logic: A graph's 2nd derivative equals zero at inflection points.

\(\displaystyle \dfrac{d^{2}y}{dx} = \sin(2x) - \cos(4x) = 0\) - How to solve from here?

 

[Bob Brown MSEE]

\(\displaystyle \dfrac{d^{2}y}{dx} = \sin(2x) - \cos(4x)\)

How many points of inflection does the graph have on the interval of \(\displaystyle [0,10]\) ?

Graph of what? Graph of y(x)?

 

[Bob Brown MSEE]

\(\displaystyle \dfrac{d^{2}y}{dx} = \sin(2x) = cos(4x)\) - How to solve from here

Typo?

 

[Jason76]

Post Edited. The graph is the graph of y(x)

 

[Deleted member 4993]

Post Edited

\(\displaystyle \dfrac{d^{2}y}{dx} = \sin(2x) - \cos(4x)\)

How many points of inflection does the graph of y(x) have on the interval of \(\displaystyle [0,10]\) ? Answer: seven

Logic: A graph's 2nd derivative equals zero at inflection points.

\(\displaystyle \dfrac{d^{2}y}{dx} = \sin(2x) - \cos(4x) = 0\) - How to solve from here?

This is what happens when algebraic/trigonometric foundation is not properly poured.

sin(2x) = cos(4x)

sin(2x) = 1 - 2sin²(2x)

Now what do you .....

 

[DrPhil]

\(\displaystyle \dfrac{d^{2}y}{dx} = \sin(2x) - \cos(4x)\)

You haven't typed the 2nd derivative correctly. Both differentiations are with respect to \(\displaystyle x\), so the denominator needs to be squared.

\(\displaystyle \dfrac{d^{2}y}{dx^2} = \dfrac{d}{dx} \left[\dfrac{d}{dx}(y)\right]\)

You have to have the same number if infinitesimals in numerator and denominator of the operator. By convention, \(\displaystyle dx^2\) means the square of \(\displaystyle dx\). If instead you wanted to write the increment of the square, you would have to use parentheses: \(\displaystyle d(x^2)\).

 

[lookagain]

Post Edited


Logic: A graph's 2nd derivative equals zero at inflection points.

Inflection points can also occur where the second derivative is undefined.

Look at x = 0 at \(\displaystyle \ y \ = \ \sqrt[3]{x}.\)

To the left of x = 0, the graph is concave up, and to the right of x = 0, the graph is concave down.

There is a point at (0, 0) in the graph of the original function, and that function is continuous everywhere.