Math Problem Statement

For what value of x does the function f(x) = e^(x^10 / 10 + 11px) have a critical point?

Solution

The problem asks for the value of xx where the function f(x)=ex10/10+11pxf(x) = e^{x^{10} / 10 + 11px} has a critical point. To find the critical point, we need to compute the derivative f(x)f'(x), set it equal to zero, and solve for xx.

Steps:

  1. Compute the first derivative f(x)f'(x): The function is f(x)=eg(x)f(x) = e^{g(x)}, where g(x)=x1010+11pxg(x) = \frac{x^{10}}{10} + 11px. Using the chain rule: f(x)=eg(x)g(x)f'(x) = e^{g(x)} \cdot g'(x) Now, find g(x)g'(x): g(x)=ddx(x1010+11px)=x9+11pg'(x) = \frac{d}{dx} \left( \frac{x^{10}}{10} + 11px \right) = x^9 + 11p

    Therefore, the derivative becomes: f(x)=ex1010+11px(x9+11p)f'(x) = e^{\frac{x^{10}}{10} + 11px} \cdot (x^9 + 11p)

  2. Set f(x)=0f'(x) = 0: Since ex1010+11px0e^{\frac{x^{10}}{10} + 11px} \neq 0 for any real xx, the critical points occur when: x9+11p=0x^9 + 11p = 0

  3. Solve for xx: x9=11px^9 = -11p x=(11p)1/9x = (-11p)^{1/9}

Thus, the function has a critical point when x=(11p)1/9x = (-11p)^{1/9}.

Would you like more details or have any further questions? Here are 5 related questions to consider:

  1. What is the definition of a critical point in calculus?
  2. How does the chain rule apply when differentiating exponential functions?
  3. Can pp be any value, and how does it affect the critical point?
  4. What does the behavior of the second derivative tell us about the nature of this critical point?
  5. How would this problem change if f(x)f(x) involved different exponents or terms?

Tip: To analyze critical points thoroughly, always check both the first and second derivatives for sign changes and concavity.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Critical Points

Formulas

f'(x) = e^g(x) * g'(x)
g'(x) = x^9 + 11p
f'(x) = 0 => x^9 + 11p = 0

Theorems

Chain Rule

Suitable Grade Level

College level (Calculus 1 or 2)