Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

\\arctan(x + t)e^{-\\frac{1}{2}t} t = 2

Question

\arctan(x + t)e^{-\frac{1}{2}t}

t = 2

Explanation:

🆕 New Concept Discovered: Function Transformations
How parameters shift and scale a parent graph.

Step 1: Identify the parent function and parameters

The given function is:

$$ f(x) = \arctan(x + t)e^{-\frac{1}{2}t} $$

We are given a specific value for the parameter \( t \):

$$ t = 2 $$

Substituting \( t = 2 \) into the function gives:

$$ f(x) = \arctan(x + 2)e^{-\frac{1}{2}(2)} $$
$$ f(x) = \arctan(x + 2)e^{-1} $$
$$ f(x) = \frac{\arctan(x + 2)}{e} $$

Step 2: Analyze the horizontal shift

The parent function is \( y = \arctan(x) \).

Replacing \( x \) with \( x + 2 \) shifts the entire graph horizontally.

  • Adding a positive constant inside the function argument, \( (x + d) \), shifts the graph to the left by \( d \) units.
  • Therefore, \( \arctan(x + 2) \) shifts the standard arctangent graph to the left by \( 2 \) units.
  • The inflection point (center) of the standard arctangent graph is at \( (0,0) \). Shifting it left by \( 2 \) units moves this center point to \( (-2, 0) \). This matches the x-intercept shown on the graph.

Step 3: Analyze the vertical scaling

The factor \( e^{-1} = \frac{1}{e} \approx 0.368 \) acts as a vertical compression factor.

  • The horizontal asymptotes of the standard \( \arctan(x) \) function are \( y = \frac{\pi}{2} \approx 1.57 \) and \( y = -\frac{\pi}{2} \approx -1.57 \).
  • Multiplying by \( e^{-1} \) scales these asymptotes vertically.
  • The new horizontal asymptotes are:
$$ y = \frac{\pi}{2e} \approx \frac{1.57}{2.718} \approx 0.58 $$
$$ y = -\frac{\pi}{2e} \approx -\frac{1.57}{2.718} \approx -0.58 $$

This matches the graph, where the curve flattens out vertically well below \( y = 1 \) and above \( y = -1 \).

Answer:

The graph represents the function \( f(x) = \arctan(x + t)e^{-\frac{1}{2}t} \) evaluated at \( t = 2 \), which simplifies to:

$$ f(x) = \frac{\arctan(x + 2)}{e} $$

This is a standard arctangent curve that has been:

  1. Shifted left by 2 units (moving the center inflection point to \( (-2, 0) \)).
  2. Vertically compressed by a factor of \( \frac{1}{e} \approx 0.368 \) (making the horizontal asymptotes \( y \approx \pm 0.58 \)).