2. A cubic function is a polynomial of degree 3 and has the form y=mx^3+bx^2+cx+d; m≠0. What is the maximum quantity of local extreme values a given cubic function can have?

a. 2
b. 1
c. 0
d. 3

Is it (a)??

2. Let f(x)=xln(x). The minimum value attained by f is
a) there is no minimum
b)1/e
c)-1/e
d)-1
e)0

is it (a) ??

#2 (a) is correct

#3 f' = lnx + 1
The only extremum is at x = 1/e
f" = 1/x, which is positive at 1/e, so f(1/e) is a minimum.

Looks like (c) is the answer

See the graph at

http://www.wolframalpha.com/input/?i=x*lnx

Ignore tha part where x<0, since lnx is not real there.

For the first question, the maximum quantity of local extreme values that a cubic function can have is 1. This is because a cubic function is a polynomial of degree 3, and it can have at most one local maximum or one local minimum point.

To determine the maximum quantity of local extreme values, you need to analyze the behavior of the cubic function by looking at its first and second derivatives.

The first derivative, f'(x), gives information about the slope of the function. To find the critical points (points where the first derivative is equal to zero or undefined), set f'(x) = 0 and solve for x. If there is more than one critical point, it means the function may have multiple local extreme values.

The second derivative, f''(x), gives information about the concavity of the function. To determine the nature of the critical points (whether they are local maxima or local minima), you can analyze the sign of the second derivative at those points. If f''(x) is positive at a critical point, it indicates a local minimum, and if f''(x) is negative, it indicates a local maximum.

In the case of a cubic function, there can be at most one critical point (corresponding to one extreme value), as confirmed by the options provided. Thus, the answer to the first question is b) 1.

Now for the second question, let's find the minimum value attained by the function f(x) = x ln(x).

To determine the minimum value, we need to find the critical points. Since the function is defined for x > 0, the natural logarithm of x is defined, and we can find the critical points by setting the derivative equal to zero.

The derivative of the function f(x) = x ln(x) can be found using the product rule as follows:

f'(x) = ln(x) + 1.

Setting f'(x) equal to zero, we get:

ln(x) + 1 = 0.

Solving for x, we find:

ln(x) = -1,
x = e^(-1).

Now, we need to check if this critical point is a minimum or not. We can do this by analyzing the behavior of the function around this point and checking the second derivative.

The second derivative, f''(x), can be found by taking the derivative of the first derivative:

f''(x) = 1/x.

Since the second derivative is always positive for x > 0, this means that the critical point x = e^(-1) is a minimum.

Therefore, the minimum value attained by the function f(x) = x ln(x) is e^(-1), which corresponds to option b) 1/e.

I hope this explanation helps! Let me know if you have any further questions.