# How to solve for x using 2nd derivative?

• MHB
• gevni
In summary: CT/(A-B))In summary, the conversation discusses finding the minimum root (x) of a formula and taking the 2nd derivative to do so. The formula for cost calculation is given and the steps for finding the minimum value are outlined. The conversation also clarifies that the value of x that makes y' = 0 and y'' > 0 is x = sqrt(CT/(A-B)).
gevni
Hi, I am trying to find the minimum root (x) of one formula. For that, I took 2nd derivative and got this equation.

$2 \times A \times (\frac{T}{x^3})=0$

Here A,T,x are greater then 0. I don't know how to proceed further, how to solve it for x? Can you please guide me?

Can you post the original problem in its entirety?

Ok! so I have a formula for cost calculation. I have to find the value of x where that formula will gives minimum value as cost should not be equal to zero, it has some minimum value.
The original formula is
$cost= A\times x+ B \times(n-x)+ C\times \frac{T}{x}$

As I am only interested in minimum value of this formula. So I took 2nd derivative of it. That results in ;

$2 \times C \times (\frac{T}{x^3})=0$

Here, A,B,C,T,n and x are greater then zero and A,B,C,T,n are known variables. I don't know how to solve it for x after 2nd derivative ?

$y = Ax + B(n-x) + \dfrac{CT}{x}$ will have a minimum at the x-value where $y' = 0$ and $y'' > 0$

$y' = 0$ at $x = \sqrt{\dfrac{CT}{A-B}}$, assuming $A > B$

$y'' = \dfrac{2CT}{x^3} > 0$, so the minimum occurs at the x-value stated above

skeeter said:
$y = Ax + B(n-x) + \dfrac{CT}{x}$ will have a minimum at the x-value where $y' = 0$ and $y'' > 0$

$y' = 0$ at $x = \sqrt{\dfrac{CT}{A-B}}$, assuming $A > B$

$y'' = \dfrac{2CT}{x^3} > 0$, so the minimum occurs at the x-value stated above

Thank you soooo much for helping. I was confused about whether I have to use
$y' = 0$ at $x = \sqrt{\dfrac{CT}{A-B}}$
or
$y' = 0$ at $x = - \sqrt{\dfrac{CT}{A-B}}$

gevni said:
Thank you soooo much for helping. I was confused about whether I have to use
$y' = 0$ at $x = \sqrt{\dfrac{CT}{A-B}}$
or
$y' = 0$ at $x = - \sqrt{\dfrac{CT}{A-B}}$

you stated ...

Here, A,B,C,T,n and x are greater then zero

positive square root, correct?

gevni said:
Hi, I am trying to find the minimum root (x) of one formula. For that, I took 2nd derivative and got this equation.

$2 \times A \times (\frac{T}{x^3})=0$

Here A,T,x are greater then 0. I don't know how to proceed further, how to solve it for x? Can you please guide me?
This is the same as $\frac{2AT}{x^3}= 0$.
But a fraction is 0 only when the numerator is 0. There is no value of x that makes it 0!

gevni said:
Ok! so I have a formula for cost calculation. I have to find the value of x where that formula will gives minimum value as cost should not be equal to zero, it has some minimum value.
The original formula is
$cost= A\times x+ B \times(n-x)+ C\times \frac{T}{x}$

As I am only interested in minimum value of this formula. So I took 2nd derivative of it. That results in ;

$2 \times C \times (\frac{T}{x^3})=0$

Here, A,B,C,T,n and x are greater then zero and A,B,C,T,n are known variables. I don't know how to solve it for x after 2nd derivative ?
A minimum for f(x) can occur where the first derivative is 0 and the second derivative is positive, not where the second derivative is 0!

skeeter said:
you stated ...
positive square root, correct?
yes

## 1. What is the purpose of using the 2nd derivative to solve for x?

The 2nd derivative is used to find the concavity of a function, which helps determine the direction of the function's curvature. This information is useful in solving for x because it can help identify the maximum or minimum points of the function.

## 2. How do I find the 2nd derivative of a function?

To find the 2nd derivative, you first need to take the derivative of the function to get the 1st derivative. Then, take the derivative of the 1st derivative to get the 2nd derivative. Alternatively, you can use the power rule or product rule to find the 2nd derivative.

## 3. Can the 2nd derivative be used to solve for x in any type of function?

Yes, the 2nd derivative can be used to solve for x in any type of function, as long as the function is continuous and differentiable. This means that the function must have a defined value at every point and its derivative must exist at every point.

## 4. What are the steps for solving for x using the 2nd derivative?

The steps for solving for x using the 2nd derivative are as follows:

1. Take the 2nd derivative of the function.
2. Set the 2nd derivative equal to 0 and solve for x.
3. Plug the value of x into the original function to find the corresponding y-value.
4. Check the concavity of the function at the identified point using the 2nd derivative test.

## 5. What is the 2nd derivative test and how is it used to solve for x?

The 2nd derivative test is a method for determining the nature of a critical point (maximum or minimum) using the 2nd derivative of a function. It states that if the 2nd derivative is positive at a critical point, the point is a minimum, and if the 2nd derivative is negative, the point is a maximum. This information can be used to solve for x by finding the critical points and evaluating the function at those points to determine the maximum or minimum values.

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