Topic 5- Curve Sketching from 5.5

Intercepts

• Y-intercept:
  • Set $t = 0$ in the function $f(t)$.
  • In this case, $f(0) = 0$, so the function passes through the origin (0, 0).
  • There are no other intercepts.

Asymptotes

• Vertical Asymptotes:
  • The function is well-defined for all values of $t$, so there are no vertical asymptotes.
  • There are no issues with division by zero.
  • The function will give a perfectly well-defined real number for any $t$.
• Horizontal Asymptotes:
  • Relate to the end behavior of the function as $t$ tends to infinity.
  • Consider the limit of the function as $t$ tends to infinity: $\lim_{t \to \infty} f(t)$.
  • This limit is in indeterminate form of type $0 \times \infty$, because as $t$ becomes very large, the factor $t$ becomes very large (tending to infinity) and the exponential term tends to zero.
  • L'Hôpital's Rule can be used to evaluate this limit.
    • Rewrite the function to apply L'Hôpital's Rule: $e^{-0.5t} = \frac{1}{e^{0.5t}}$.
    • Now the limit is of type $\frac{\infty}{\infty}$.
    • Apply L'Hôpital's Rule by differentiating the numerator and denominator:
      $\lim<em>{t \to \infty} \frac{10t}{e^{0.5t}} = \lim</em>{t \to \infty} \frac{10}{0.5e^{0.5t}}$.
    • As $t$ tends to infinity, the denominator tends to infinity, so the limit is zero.
    • This indicates a horizontal asymptote at $y = 0$.

Critical Points

• To find critical points, differentiate the function $f(t)$ and find where the derivative is equal to zero.
• The function, $f(t) = 10te^{-0.5t}$, is a product, so use the product rule to differentiate it.
  • $f'(t) = 10e^{-0.5t} + 10t(-0.5e^{-0.5t})$.
  • Simplify the expression by taking out common factor:
  • $f'(t) = 5e^{-0.5t}(2 - t)$.
• Set the first derivative equal to zero to find critical points.
  • Since exponentials are never zero, the term in the brackets must be zero. Thus,
  • $2 - t = 0 \Rightarrow t = 2$.
  • $t = 2$ is the only critical point.
• Use the first derivative test to determine if the critical point is a local maximum or minimum.
• Consider the intervals from zero to two and from two to infinity.
• Pick test points in each interval and evaluate $f'(t)$.
  • If f'(t) > 0, the function is increasing.
  • If f'(t) < 0, the function is decreasing.
• It turns out that $f'(t)$ is positive for $t < 2$ and negative for $t > 2$.
  • Therefore, the function is increasing on the interval $[0, 2)$ and decreasing on the interval $(2, \infty)$.
  • Thus, the critical point at $t = 2$ is a local maximum or relative maximum.
• Evaluate the function at the critical point to help with graph sketching.
  • $f(2) = 10(2)e^{-0.5(2)} = 20e^{-1} = \frac{20}{e} \approx 7.4$.
  • So there is a relative maximum at $t = 2$, and $y \approx 7.4$.

Second Derivative

• To find the second derivative, differentiate the first derivative.
• $f''(t) = -5e^{-0.5t} - 5(-0.5)te^{-0.5t} + 5(-0.5)e^{-0.5t}$
• Simplify the expression by factoring out the exponential term:
  • $f''(t) = e^{-0.5t}(\frac{5}{2}t - 10)$
• To find points of inflection, set the second derivative equal to zero and solve for t.
  • Since the exponential term can never be zero, the factor in the brackets must be zero.
    • $\frac{5}{2}t - 10 = 0 \Rightarrow t = 4$.

Concavity

• Find the intervals where the second derivative is positive or negative.
• Consider the intervals from zero to four and from four to infinity.
• Pick test points in each interval and evaluate $f''(t)$.
  • If f''(t) < 0, the function is concave down.
  • If f''(t) > 0, the function is concave up.
• For $t < 4$, $f''(t) < 0$ and for $t > 4$, f''(t) > 0.
  • Therefore, the function is concave down on the interval $[0, 4)$ and concave up on the interval $(4, \infty)$.
  • So, there is a point of inflection at $t = 4$.

Graph Sketching

• Intercept at the origin (0, 0).
• Horizontal asymptote at $y = 0$ as $t$ tends to infinity.
• Relative maximum at $(2, 7.4)$.
• Point of inflection at $t = 4$, where the function changes from concave down to concave up.
• The function increases from the origin up to the relative maximum, then decreases towards the horizontal asymptote.

Symmetry

• Check if the function is even or odd.
• Substitute $-x$ into the function and see if it is equal to $f(x)$ (even) or $-f(x)$ (odd).
• In this case,
  $<br /> f(-x) = e^{-\frac{(-x)^2}{2}} = e^{-\frac{x^2}{2}} = f(x)$
• The function is even.

Intercepts

• To look for intercepts, let's start off by setting $y = 0$.
• We'd require $e^{-\frac{x^2}{2}} = 0$.
• There's no solution to this equation, so there's no x intercept.
  This function never touches the x axis.
• To find y intercepts, set $x = 0$ and substitute that into our function.
  The result, is $y=1$

Asymptotes

• Vertical Asymptotes:

  • Function has no vertical asymptotes for the same reason as in the previous example; it is defined for all values of x.

• Horizontal Asymptote

  • To find horizontal asymptote, we need to think about this limit: $\lim_{x \rightarrow \infty} e^{-\frac{x^2}{2}}$.
  • This limit is zero.
  • A horizontal Asymptote presents at $y = 0$
  • The same is true if I let $x \rightarrow -\infty$,

Critical Points

• To look for critical points, we need to work out the derivative.
• $f'(x) = -xe^{\frac{-x^2}{2}}$.
• The critical point is when $x = 0$.

First Derivative Test

• If $x$ is positive $f'(x)$ is negative.
• If $x$ is negative $f'(x)$ is positive.
• The critical point at $x = 0$ must be a relative maximum

Graph Sketching

• The intercept is at $(0, 1)$.
• The function has a relative maximum at the same point.
• The critical point, slope is zero.
• There are inflection points, mirror image of each other.