What Is the Minimum Number of Real Roots for the Given Equation?

[utkarshakash]

Homework Statement

Let f(x) be a non-constant twice differentiable function defined on R such that f(x)=f(1-x) and f'(1/4) =0 then what is the minimum number of real roots of the equation (f"(x))^2+f'(x)f'''(x)=0.

The Attempt at a Solution

f'(x)=-f'(1-x)
f"(x)=f"(1-x)
f'''(x)=-f'''(1-x)

putting x = 1/4 in the first eqn gives f'(3/4)=0
putting x=1/4 in the given equation (f"(x))^2+f'(x)f'''(x)=0 gives f"(1/4)=0. ∴f"(3/4)=0.
So I have got a total of 2 roots. But there are some more which I can't find.

 

[Saitama]

Homework Statement

Let f(x) be a non-constant twice differentiable function defined on R such that f(x)=f(1-x) and f'(1/4) =0 then what is the minimum number of real roots of the equation (f"(x))^2+f'(x)f'''(x)=0.

The Attempt at a Solution

f'(x)=-f'(1-x)
f"(x)=f"(1-x)
f'''(x)=-f'''(1-x)

putting x = 1/4 in the first eqn gives f'(3/4)=0
putting x=1/4 in the given equation (f"(x))^2+f'(x)f'''(x)=0 gives f"(1/4)=0. ∴f"(3/4)=0.
So I have got a total of 2 roots. But there are some more which I can't find.

Observe that the given equation is
$$\frac{d}{dx}(f'(x)\cdot f''(x))=0$$
See if that helps.

 

[utkarshakash]

Observe that the given equation is
$$\frac{d}{dx}(f'(x)\cdot f''(x))=0$$
See if that helps.

Let y= f'(x)f"(x)

According to Rolle's Theorem there must lie a root of dy/dx between 1/4 and 3/4. But what about other roots?

 

[Saitama]

Let y= f'(x)f"(x)

According to Rolle's Theorem there must lie a root of dy/dx between 1/4 and 3/4. But what about other roots?

I am unsure how to proceed but how about x=1/2?

At how many points is f'(x) zero? At how many points is f''(x) zero?

 

[pasmith]

Let y= f'(x)f"(x)

According to Rolle's Theorem there must lie a root of dy/dx between 1/4 and 3/4. But what about other roots?

You need to find the known zeroes of [itex]y = f'f''[/itex], at which point you can apply Rolle's theorem to get the known zeroes of [itex]y'[/itex], which is what you're interested in.

So far you know that [itex]f'(\frac14) = f'(\frac34) = 0[/itex].

What, in view of the condition [itex]f(x) = f(1-x)[/itex], can you say about [itex]f'(\frac12)[/itex]?

Now apply Rolle's theorem to [itex]f'[/itex].

 

[utkarshakash]

You need to find the known zeroes of [itex]y = f'f''[/itex], at which point you can apply Rolle's theorem to get the known zeroes of [itex]y'[/itex], which is what you're interested in.

So far you know that [itex]f'(\frac14) = f'(\frac34) = 0[/itex].

What, in view of the condition [itex]f(x) = f(1-x)[/itex], can you say about [itex]f'(\frac12)[/itex]?

Now apply Rolle's theorem to [itex]f'[/itex].

I can say that there must lie a root of y' between 1/4 and 1/2 as well as 1/2 and 3/4.

 

[Saitama]

I can say that there must lie a root of y' between 1/4 and 1/2 as well as 1/2 and 3/4.

y is also zero at some more points.

f'(x) is zero at 1/2 and 1/4 so there must be at least one point in between them where f''(x) is zero. Can you proceed now?