# Minimum number of real roots

• utkarshakash
In summary, we are trying to find the minimum number of real roots for the equation (f"(x))^2+f'(x)f'''(x)=0, given that f(x)=f(1-x) and f'(1/4)=0. By applying Rolle's theorem and considering the conditions f(x)=f(1-x) and f'(1/4)=0, we can determine that there are a minimum of 3 real roots for this equation.

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## 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.

utkarshakash said:

## 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.

Pranav-Arora said:
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?

utkarshakash said:
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?

Last edited:
utkarshakash said:
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 $y = f'f''$, at which point you can apply Rolle's theorem to get the known zeroes of $y'$, which is what you're interested in.

So far you know that $f'(\frac14) = f'(\frac34) = 0$.

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

Now apply Rolle's theorem to $f'$.

pasmith said:
You need to find the known zeroes of $y = f'f''$, at which point you can apply Rolle's theorem to get the known zeroes of $y'$, which is what you're interested in.

So far you know that $f'(\frac14) = f'(\frac34) = 0$.

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

Now apply Rolle's theorem to $f'$.

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.

utkarshakash said:
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?

## What is the "Minimum number of real roots"?

The minimum number of real roots refers to the smallest number of real solutions that a polynomial equation can have. These solutions are the values of the variable that make the equation true when plugged in.

## How can I determine the minimum number of real roots?

The minimum number of real roots can be determined by analyzing the degree of the polynomial equation. The degree is the highest exponent of the variable in the equation. For example, a polynomial equation with degree 2 can have a minimum of 2 real roots.

## What is the difference between real roots and complex roots?

Real roots are values of the variable that result in a real number solution, while complex roots are values of the variable that result in an imaginary number solution. Imaginary numbers involve the square root of -1 and are denoted by the letter "i".

## Can a polynomial equation have more than one minimum number of real roots?

No, a polynomial equation can only have one minimum number of real roots based on its degree. However, it is possible for a polynomial equation to have multiple real roots, but these roots will always be equal to or less than the minimum number of real roots.

## What is the significance of knowing the minimum number of real roots?

Knowing the minimum number of real roots can help determine the behavior of a polynomial equation. For example, if a polynomial equation has a minimum number of real roots of 2, it means that the graph of the equation will intersect the x-axis at least twice. This can be useful in understanding the overall shape and behavior of the equation's graph.