Math Problem Statement

For what value of x does the function f(x) = e^(x^4/4) + 10px have a critical point? Answer with an expression that includes p.

Solution

We are given the function:

f(x)=ex4/4+10pxf(x) = e^{x^4/4} + 10px

and asked to find for what value of xx the function f(x)f(x) has a critical point, and express the result in terms of pp.

Steps to Solve:

  1. Critical points occur where the first derivative of the function f(x)f(x) equals zero.

    We need to compute f(x)f'(x).

  2. Derivative of the given function:

    • The first term, ex4/4e^{x^4/4}, will be differentiated using the chain rule.
    • The second term, 10px10px, is linear, and its derivative is straightforward.

f(x)=ddx(ex4/4)+ddx(10px)f'(x) = \frac{d}{dx} \left( e^{x^4/4} \right) + \frac{d}{dx} \left( 10px \right)

For the first term: ddx(ex4/4)=ex4/4ddx(x44)=ex4/4x3\frac{d}{dx} \left( e^{x^4/4} \right) = e^{x^4/4} \cdot \frac{d}{dx} \left( \frac{x^4}{4} \right) = e^{x^4/4} \cdot x^3

For the second term: ddx(10px)=10p\frac{d}{dx} \left( 10px \right) = 10p

Thus, the derivative of f(x)f(x) is: f(x)=ex4/4x3+10pf'(x) = e^{x^4/4} \cdot x^3 + 10p

  1. Set the derivative equal to zero to find critical points: ex4/4x3+10p=0e^{x^4/4} \cdot x^3 + 10p = 0

  2. Solve for xx: ex4/4x3=10pe^{x^4/4} \cdot x^3 = -10p Since ex4/4e^{x^4/4} is always positive, it follows that the sign of x3x^3 must match the sign of 10p-10p. Dividing both sides by ex4/4e^{x^4/4}, we get: x3=10pex4/4x^3 = \frac{-10p}{e^{x^4/4}}

Thus, the critical points are determined by the above equation. We cannot explicitly solve this for xx without further information about pp, but this equation provides the implicit relationship between xx and pp.


Would you like more details on how to interpret this or help with other steps?

Related Questions:

  1. How do we determine if this critical point is a maximum, minimum, or saddle point?
  2. Can we further approximate xx for specific values of pp?
  3. What happens to the critical points when p=0p = 0?
  4. How does the nature of ex4/4e^{x^4/4} affect the solution for xx?
  5. What is the second derivative of f(x)f(x), and how does it help in identifying the concavity?

Tip: The nature of critical points depends not only on the first derivative but also on the second derivative (the concavity test).

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Critical Points

Formulas

f'(x) = e^(x^4/4) * x^3 + 10p
Critical points occur where f'(x) = 0

Theorems

Chain Rule in Differentiation

Suitable Grade Level

Undergraduate Calculus or Advanced High School