Let f(x) = e^sin(x^2 - ax - 6a^2), where a is a positive real number. Without computing f', show...

Question:

Let {eq}f(x) = e^{sin(x^2 - ax - 6a^2)} {/eq}, where a is a positive real number.

Without computing {eq}f' {/eq}, show that {eq}f {/eq} has a critical point.

Critical Points:

The critical points are the extreme points of a function at which function, f(x) attains its maximum and minimum value. In this question, we will calculate the upper endpoint and the lower endpoint of the function in which the function will be bounded.

Answer and Explanation:

Given that,

{eq}f(x) = e^{sin(x^2-ax-6a^2)} {/eq}

Let us consider,

{eq}y(x) = x^2-ax-6a^2 {/eq}

{eq}y(x) {/eq} is bounded.

{eq}-1 \le sin(y(x)) \le 1 {/eq}


Now,

{eq}e^{-1} \le e^{sin(y(x))} \le e^{1} {/eq}


As f(x) is differential function any point at which it attains. Its absolute maximum value will be a critical point.

The absolute maximum value {eq}=e. {/eq}

The absolute minimum value {eq}=e^{-1} {/eq}


Learn more about this topic:

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Finding Critical Points in Calculus: Function & Graph

from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9
195K

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