QA: Sketch a function that behaves as described (separately for i,ii); also, on the same axes, sketch the derivative of that function.
i) increasing on [0,5], f ‘(5)=0, then decreasing on [5,infinity)
ii) increasing on [0,5], derivative = “infinity” at x=5, and increasing on (5,infinity)
QB: Consider f(x)=sin(x)+x/2 (note that this is different than (sin(x)+x)/2 ). Would you say that f is increasing on [0,20] (try plotting it)? Would most people who haven’t taken a calc class say that f is increasing on [0,20] ? Explain.
QC: Sketch a continuous function on [0,10] that has a global minimum at x=3, a local minimum at x=8, and a global max (on the interval) at x=10. Can you do it without having a local max in the interval?
QD: (i) Find all critical numbers of exp(-x)*x^3 on [0,infinity]. Note that the x^3 is not in the exponent of the “e”: it’s not exp(-x*x^3). While I usually avoid problems where you have to factor things, factoring will help a lot on this problem. Also, you might want the logic that if a*b*c=0, then at least one of a, b, or c must be 0. This shape is related to a “Gamma distribution” in probability/statistics.
(ii) Now find all the critical numbers of exp(-x^10)*x^9 on [0,infinity]. Again, note that the x^9 is not part of the exponent. This shape is related to a “Weibull distribution” in probability/statistics.
(iii) What do you notice about the shape of the two curves above, compared to each other (ignoring their size and spread)?
QE: Find the local minimum of f(x)=x - k*Ln(x), for constant k>0. This is related to “interior point methods” in optimization algorithms (which I teach about in Math 560). It’s a way of saying “minimize f(x)=x for x>=0” without having a separate constraint like “x>=0”; it turns a minimum-is-at-an-endpoint kind of problem into a minimum-is-in-the-interior-of-the-interval problem. This is handy if you have more than one dimension and finding the cornerpoints would take too long.On the other hand, it turns a linear problem (with constraints) into a nonlinear problem, which for decades seemed like a bad trade-off to algorithm designers, but turns out it can be a good idea.
another homework dump?
I'll get you started.
A: Sketch a function that behaves as described (separately for i,ii); also, on the same axes, sketch the derivative of that function.
i) increasing on [0,5], f '(5)=0, then decreasing on [5,∞)
ii) increasing on [0,5], f(5) = ∞, and increasing on (5,∞)
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i) so, basically bell-shaped, peaking at x=5. Now consider 1/(x^2+1) which is bell-shaped, but peaks at x=0. Shifting it 5 to the right, that would be
f(x) = 1/((x-5)^2 + 1)
ii) clearly there is an asymptote at x=5. Consider -1/x, which has its asymptote at x=0. Shift that right by 5, and you have
f(x) = -1/(x-5) = 1/(5-x)
Now go to your favorite online online graphing site and check the graphs.
See
https://www.wolframalpha.com/input/?i=plot+y%3D1%2F%28%28x-5%29%5E2%2B1%29%2C+y%3D1%2F%285-x%29
What do you have to say on the other parts?
A: i) To sketch a function that behaves as described, we can start with a straight line that increases from 0 to 5 and then decreases from 5 to infinity. At x = 5, the derivative of the function is 0, which means the slope of the line is 0 at that point.
ii) For the second part, we can start with a straight line that increases from 0 to 5. At x = 5, the derivative "infinity" indicates a vertical tangent line. After x = 5, the derivative remains positive, indicating that the function is increasing.
On the same axes, we can sketch the derivative of the function as a separate curve. For part i), the derivative is positive from 0 to 5, and then becomes negative after x = 5. For part ii), the derivative is positive from 0 to 5, and continues to be positive after x = 5.
B: To determine if f(x) = sin(x) + x/2 is increasing on [0,20], we can plot the function or analyze its behavior. Based on the function, it is a combination of a sinusoidal function and a linear function. The sine component oscillates between -1 and 1, but the linear component (x/2) increases as x increases. Overall, the function will increase, but with some oscillations. Most people who haven't taken a calculus class may not have the tools to analyze the behavior of the function, so their perception may vary.
C: To sketch a continuous function on [0,10] with a global minimum at x = 3, a local minimum at x = 8, and a global maximum at x = 10, we can draw a curve that initially increases, reaches a local minimum at x = 8, and then increases again to a global maximum at x = 10. It is possible to have a function with only a local maximum in the interval, as the local maximum does not necessarily need to exist in this case.
D:
i) To find the critical numbers of exp(-x) * x^3 on [0, infinity], we need to find the values of x where the derivative is either zero or undefined. Taking the derivative of the function, we have f'(x) = exp(-x) * (3x^2 - x^3).
Setting f'(x) = 0 and solving for x, we get:
exp(-x) * (3x^2 - x^3) = 0
Using the factoring logic mentioned, we can conclude that either exp(-x) = 0 or 3x^2 - x^3 = 0. However, exp(-x) is never zero, so we only consider the second equation. Simplifying further, we have:
3x^2 - x^3 = x^2(3 - x) = 0
Setting each factor equal to zero:
x^2 = 0 or (3 - x) = 0
x = 0 or x = 3
Therefore, the critical numbers are x = 0 and x = 3.
ii) To find the critical numbers of exp(-x^10) * x^9 on [0, infinity], we follow a similar process as in part i). Taking the derivative, we have f'(x) = exp(-x^10) * (9x^8 - 10x^9).
Setting f'(x) = 0, we get:
exp(-x^10) * (9x^8 - 10x^9) = 0
Again, using the factoring logic, we conclude that either exp(-x^10) = 0 or 9x^8 - 10x^9 = 0. However, exp(-x^10) is never zero, so we only consider the second equation. Simplifying, we have:
9x^8 - 10x^9 = x^8(9 - 10x) = 0
Setting each factor equal to zero:
x^8 = 0 or (9 - 10x) = 0
x = 0 or x = 9/10
Therefore, the critical numbers are x = 0 and x = 9/10.
iii) Comparing the shape of the two curves, we notice that as the exponent in the function increases, the curve becomes steeper and increasingly concentrated around the y-axis. This results in the function having more critical points closer to the origin, and an overall narrower spread.
E: To find the local minimum of f(x) = x - k * Ln(x), where k > 0, we need to find the value of x where the derivative is zero. Taking the derivative of the function, we have f'(x) = 1 - k/x.
Setting f'(x) = 0 and solving for x, we get:
1 - k/x = 0
Simplifying, we have:
k/x = 1
Cross-multiplying, we get:
k = x
Therefore, the local minimum occurs when x = k.
QA:
The first step to sketching a function that behaves as described in part (i) is to understand the given information. The function is increasing on the interval [0,5], has a derivative of 0 at x=5, and then it decreases on the interval [5,∞).
To sketch the function, start by drawing a horizontal line segment from (0,0) to (5,K), where K is any positive value. This represents the function increasing on [0,5]. Then, draw a horizontal line segment from (5,K) to the right side of the graph, decreasing indefinitely. This represents the function decreasing on [5,∞).
Next, to sketch the derivative of the function, we need to understand the behavior of the derivative. The derivative is 0 at x=5.
To represent this, draw a horizontal line segment at y=0 across the entire graph. This represents the derivative being 0 at x=5.
QB:
To determine if the function f(x)=sin(x)+x/2 is increasing on the interval [0,20], we can analyze the function and its derivative.
Plot the function f(x) = sin(x) + x/2 on the given interval. You can use a graphing calculator or software to plot the graph.
Now, to determine if the function is increasing, we need to check if its derivative is positive or non-negative on the interval [0,20]. The derivative of f(x) is determined by taking the derivative of sin(x) and x/2 separately.
Taking the derivative of sin(x) gives us cos(x), and the derivative of x/2 is 1/2.
Now, plot the derivative function f'(x) = cos(x) + 1/2 on the same graph as the original function.
Look at the graph and observe the values of the derivative function. If the derivative is positive or non-negative on the interval [0,20], it indicates that the original function is increasing. If there are any negative portions of the derivative function, it means that the original function is decreasing.
Now, consider if most people who haven't taken a calculus class would say that f is increasing on [0,20].
Based on the form of the function f(x) = sin(x) + x/2, where it consists of both a trigonometric function and a linear function, it is unlikely that people without calculus knowledge would be able to determine the increasing or decreasing behavior of the function on the interval [0,20]. Calculus concepts are required to analyze these types of functions in detail and determine their increasing or decreasing nature.
QC:
To sketch a continuous function on the interval [0,10] that has a global minimum at x=3, a local minimum at x=8, and a global maximum at x=10, follow these steps:
Start by drawing a curved line that slopes downward from the left side of the graph to x=3. This represents the function's global minimum at x=3.
Next, draw a curved line that slopes upward from x=3 to x=8. This represents the function's local minimum at x=8.
Finally, draw a curved line that slopes downward from x=8 to the right side of the graph, representing the function's global maximum at x=10.
It is possible to achieve this shape without having a local maximum within the interval [0,10]. The function can continuously decrease from the local minimum at x=8 to the global maximum at x=10.
QD:
(i) To find all critical numbers of the function exp(-x)*x^3 on the interval [0,∞), we need to find the values of x where the derivative of the function is either zero or undefined.
First, find the derivative of the function. The derivative of exp(-x)*x^3 can be found using the product rule or by applying the chain rule. The derivative is given by: d/dx(exp(-x)*x^3) = -exp(-x)*x^3 + 3exp(-x)*x^2.
Next, set the derivative equal to zero and solve for x to find where it is equal to zero. Solve the equation -exp(-x)*x^3 + 3exp(-x)*x^2 = 0.
Factor out the common term exp(-x), giving you x^3 - 3x^2 = 0.
Factor the equation further: x^2(x-3) = 0.
Now, set each factor equal to zero and solve for x. The solutions are x=0 and x=3.
Therefore, the critical numbers of exp(-x)*x^3 on [0,∞) are x=0 and x=3.
(ii) Similarly, find the derivative of exp(-x^10)*x^9: d/dx(exp(-x^10)*x^9) = -10x^9exp(-x^10)*x^9 + 9exp(-x^10)*x^8.
Set the derivative equal to zero: -10x^9exp(-x^10)*x^9 + 9exp(-x^10)*x^8 = 0.
Factor out the common term exp(-x^10), giving you x^9 - 9x^8 = 0.
Factor the equation further: x^8(x-9) = 0.
Set each factor equal to zero and solve for x. The solutions are x=0 and x=9.
Therefore, the critical numbers of exp(-x^10)*x^9 on [0,∞) are x=0 and x=9.
(iii) One noticeable difference between the shapes of the two curves is their behavior as x approaches infinity. The first curve, exp(-x)*x^3, approaches zero as x approaches infinity, indicating a decay or decline. In contrast, the second curve, exp(-x^10)*x^9, approaches infinity as x approaches infinity, indicating growth or increase.