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.

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{t \to \infty} \frac{10t}{e^{0.5t}} = \lim{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.

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.

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.

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.

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.

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,

f(-x) = e^{-\frac{(-x)^2}{2}} = e^{-\frac{x^2}{2}} = f(x)
The function is even.

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

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,