Hi,

I am lost on couple of problems and i was wondering if you could just help me out with them.

1. Find and classify the points of discontinuity of the function F(x) = (x^2+7x+12)/(x^3-9x)

now for this problem i know there is going to be an infinite discontinuity because of the asymptotes, but at what points? is it going to be at x=3 x=-3 x=0 OR since after factoring top and bottom the result is (x+4)/x(x-3) so is it going to be just x=0 x=3

Also, how does canceling after factoring affect the graph; does that mean there is going to be a hole (and a removable discontinuity)? I kind of forgot this from precalculus.

2. Find all points where the tangent line to y=x^3-6x+12 has slope -1.

-lost on this one

3. Use the table below, which shows values of f(x) for x near 2, to find the slope of a secant line that is an estimate for f '(2)
x 1.8 1.9 2.0 2.1 2.2
f(x) 2.24 2.27 2.30 2.33 2.37

4. F(x) = x^2 if x<1
4-kx if x >or equal to 1

For the value of k, is f(x) differentiable at x=1? explain your answer

got 3 for the value of k

Your chances of getting questions answered promptly, or at all, greatly improve if you post them one at a time. Many of the volunteers here do no have time to answer four questions of this complexity in one sitting.

Let's consider question #2:

The slope to the tangent line of a function f(x) is f'(x) = df/dx

Thus you want to know where
f'(x) = 3x^2 -6 = -1
3x^2 = 5
x = +or- sqrt(5/3)
= +1.291 or -1.291

Sure, I can help you with these problems. Let's go through each one step by step.

1. To find and classify the points of discontinuity of the function F(x) = (x^2+7x+12)/(x^3-9x), we need to identify any values of x where the function is not defined or has a vertical asymptote.

First, let's consider the denominator and set it equal to zero: x^3 - 9x = 0. This can be factored as x(x^2 - 9) = 0, which gives us x = 0 and x = ±3 as potential points of discontinuity.

Next, let's examine the simplified function after factoring the numerator and denominator: F(x) = (x+4)/(x(x-3)). This means that the function is not defined at x = 0 and x = 3 because they make the denominator zero. Hence, there are points of discontinuity at x = 0 and x = 3.

Upon canceling common factors, any factor that is common to both the numerator and the denominator is eliminated, resulting in a hole in the graph. In this case, after canceling (x+4), we are left with the simplified function F(x) = 1/(x-3) for x ≠ 0. Since x = 0 is a point of discontinuity, we can say that the graph has a removable discontinuity (a hole) at x = 0. It does not have an effect on the vertical asymptotes.

2. To find all points where the tangent line to y = x^3 - 6x + 12 has a slope of -1, we need to find the x-values that will satisfy the given condition.

The slope of a function can be found by taking the derivative of the function with respect to x. In this case, the derivative of y = x^3 - 6x + 12 is y' = 3x^2 - 6.

Setting this derivative equal to -1, we have 3x^2 - 6 = -1. Solving this equation will give us the x-values that satisfy the condition.

3x^2 - 6 = -1
3x^2 = 5
x^2 = 5/3
x = ±sqrt(5/3)

So, the tangent line to y = x^3 - 6x + 12 has a slope of -1 at x = -sqrt(5/3) and x = sqrt(5/3).

3. To find the slope of a secant line that is an estimate for f'(2) using the table of values, we need to consider two points on the graph that are close to x = 2.

Looking at the table, we see that x = 2.0 has a corresponding f(x) value of 2.30. The closest points on both sides of x = 2 are x = 1.9 and x = 2.1, which have f(x) values of 2.27 and 2.33, respectively.

The slope of the secant line between two points (x1, f(x1)) and (x2, f(x2)) is given by (f(x2) - f(x1))/(x2 - x1). In this case, (f(x2) - f(x1))/(x2 - x1) = (2.33 - 2.27)/(2.1 - 1.9) = 0.06/0.2 = 0.3.

Therefore, the slope of the secant line that is an estimate for f'(2) is 0.3.

4. Regarding the value of k for which the function F(x) = x^2 if x < 1, and F(x) = 4 - kx if x ≥ 1 is differentiable at x = 1, we need to check if the function is continuous at x = 1 and if the left and right derivatives exist and are equal.

For F(x) to be continuous at x = 1, the values of F(x) from the left and right sides of x = 1 should be equal. Plugging in x = 1 into both expressions, we have:

For x < 1: F(1) = 1^2 = 1
For x ≥ 1: F(1) = 4 - k(1) = 4 - k

To be differentiable at x = 1, the left and right derivatives should exist and be equal. The derivative of F(x) = x^2 is F'(x) = 2x, and the derivative of F(x) = 4 - kx is F'(x) = -k.

Taking the derivative from the left side of x = 1: F'(1-) = 2(1) = 2
Taking the derivative from the right side of x = 1: F'(1+) = -k

For F(x) to be differentiable at x = 1, F'(1-) should be equal to F'(1+), so 2 = -k.

Therefore, the value of k for which F(x) is differentiable at x = 1 is k = -2.

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