Question

Find the point of inflection of y=2sin(x)+cos(2x) ;[0,pi]

Answer 1

I already found the second derivative which is
\[-2\sin(x)-4\cos(2x)\]

Answer 2

I am just having trouble finding when is it 0

Answer 3

Any \(x=\alpha\) that satisfies \(f''(\alpha)=0\), as long as the signs on the sides of \(f''(\alpha)\) are opposite (so that for a sufficiently small \(\delta\), you have \(f''(\alpha-\delta)\) and \(f''(\alpha+\delta)\) with opposite signs).

Answer 4

I will say this in a better less abstract term.

Answer 5

Just set the 2nd derivative = to 0 and solve the resulting equation for x.
Can you do that?

Answer 6

Suppose that you have a function \(f\), and \(f''(3)=0\), then \(f''(3.04)\) and \(f''(2.96)\) have opposite signs.

Answer 7

(If the signs of \(f''(3.04)\) and \(f''(2.96)\) are not opposite, - i.e. both negative or both positive, then \(x=3\) is not an inflection point.)

Answer 8

So, at first you have to find the \(x\) values, (the set of \(x\) values) that satisfies \(f''(x)=0\).

Answer 9

\(-2\sin(x)-4\cos(2x)=0\)
\(\sin(x)=-2(1-2\sin^2(x))\)
\(4(\sin x)^2-\sin(x)-2=0\)

proceed with a substitution

Answer 10

(or with questions if you have them)

Answer 11

I am lost on how you got sinx on one side by itself @SolomonZelman

Answer 12

\(-2\sin(x)-4\cos(2x)=0\)
\(\color{red}{-2\sin(x)=4\cos(2x)}\)
\(\color{red}{\sin(x)=-2\cos(2x)}\)
\(\sin(x)=-2(1-2\sin^2(x))\)
\(\color{red}{\sin(x)=-2+4\sin^2(x)}\)
\(4(\sin x)^2-\sin(x)-2=0\)

the steps in red are the once I omitted
(due to thinking they are obvious)

Answer 13

hmmm

Answer 14

I understand how you divided by 2

Answer 15

-2

Answer 16

you don't possess a firm knowledge of algebra?

Answer 17

Its been quite a while

Answer 18

Algebra, as a matter of fact, is applied constantly to any math problem ...
(Well, maybe that is just my experience alone)

Answer 19

In any case, can you solve the equation from there?

Answer 20

so would it also be cos(2x)=sin(x)

Answer 21

sin(x)=−2cos(2x)

Answer 22

I have already done the algebra up to some point ...
sub z=sin(x)

Answer 23

and solve the quadratic equation.