Math Problem Statement

Consider f(x) = e^x sin x on the interval [-π, π]. You may use, without having to prove it, that f'(x) = e^x (sin x + cos x) and f''(x) = 2e^x cos x. (a) Give all intervals (within the given domain) where f(x) is increasing. (b) Give all intervals (within the given domain) where f(x) is concave up.

Solution

Let's carefully solve each part of the question.

Given:

  1. f(x)=exsinxf(x) = e^x \sin x on the interval [π,π][-\pi, \pi].
  2. The derivatives provided are:
    • f(x)=ex(sinx+cosx)f'(x) = e^x (\sin x + \cos x)
    • f(x)=2excosxf''(x) = 2e^x \cos x.

Part (a): Where is f(x)f(x) increasing?

To determine where f(x)f(x) is increasing, we need to analyze where f(x)>0f'(x) > 0 (the first derivative is positive).

From the given derivative: f(x)=ex(sinx+cosx).f'(x) = e^x (\sin x + \cos x).

Steps:

  1. Factor exe^x: Since ex>0e^x > 0 for all real xx, the sign of f(x)f'(x) depends only on sinx+cosx\sin x + \cos x.

  2. Solve sinx+cosx>0\sin x + \cos x > 0: To solve this inequality, notice: sinx+cosx=2sin(x+π4).\sin x + \cos x = \sqrt{2} \sin \left( x + \frac{\pi}{4} \right). The sign of sin(x+π4)\sin \left( x + \frac{\pi}{4} \right) determines where sinx+cosx\sin x + \cos x is positive.

    • sinθ>0\sin \theta > 0 when θ(0,π)\theta \in (0, \pi) within a 2π2\pi-period.

    Therefore: x+π4(0,π)    x(π4,3π4).x + \frac{\pi}{4} \in (0, \pi) \implies x \in \left( -\frac{\pi}{4}, \frac{3\pi}{4} \right).

  3. Final Answer for Part (a): f(x)f(x) is increasing on the interval: (π4,3π4).\left( -\frac{\pi}{4}, \frac{3\pi}{4} \right).


Part (b): Where is f(x)f(x) concave up?

To determine where f(x)f(x) is concave up, we analyze where f(x)>0f''(x) > 0 (the second derivative is positive).

From the given second derivative: f(x)=2excosx.f''(x) = 2e^x \cos x.

Steps:

  1. Factor 2ex2e^x: Since 2ex>02e^x > 0 for all xx, the sign of f(x)f''(x) depends only on cosx\cos x.

  2. Solve cosx>0\cos x > 0:

    • cosx>0\cos x > 0 when x(π2,π2)x \in (-\frac{\pi}{2}, \frac{\pi}{2}) within the interval [π,π][-\pi, \pi].
  3. Final Answer for Part (b): f(x)f(x) is concave up on the interval: (π2,π2).\left( -\frac{\pi}{2}, \frac{\pi}{2} \right).


Final Results:

  1. (a) f(x)f(x) is increasing on (π4,3π4)\left( -\frac{\pi}{4}, \frac{3\pi}{4} \right).
  2. (b) f(x)f(x) is concave up on (π2,π2)\left( -\frac{\pi}{2}, \frac{\pi}{2} \right).

Do you need further clarification or a detailed graph of these intervals?

Here are 5 related questions you might consider:

  1. How do you determine the intervals where f(x)f(x) is decreasing?
  2. What is the behavior of f(x)f(x) at the critical points?
  3. How do you graph f(x)=exsinxf(x) = e^x \sin x and identify its key features?
  4. How can you use the first derivative test to confirm local maxima or minima?
  5. How is concavity related to the inflection points of f(x)f(x)?

Tip: To determine where a function is increasing, always analyze the sign of the first derivative!

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Increasing and Decreasing Functions
Concavity

Formulas

First Derivative: f'(x) = e^x (sin x + cos x)
Second Derivative: f''(x) = 2e^x cos x

Theorems

A function is increasing when f'(x) > 0.
A function is concave up when f''(x) > 0.

Suitable Grade Level

Grades 11-12 (Advanced Calculus)