Determine whether Rolle's Theorem can be applied to f on the closed interval
[a, b].
f(x) = −x2 + 3x, [0, 3]
Yes, Rolle's Theorem can be applied.No, because f is not continuous on the closed interval [a, b].No, because f is not differentiable in the open interval (a, b).No, because f(a) ≠ f(b).
If Rolle's Theorem can be applied, find all values of c in the open interval
(a, b)
such that
f '(c) = 0.
(Enter your answers as a comma-separated list. If Rolle's Theorem cannot be applied, enter NA.)
c =

Answers

Answer 1

Answer:

Yes, Rolle's theorem can be applied

There is only one value of c such that f'(c) = 0, and this is c = 1.5 (or 3/2 in fraction form)

Step-by-step explanation:

Yes, Rolle's theorem can be applied on this function because the function is continuous in the closed interval (it is a polynomial function) and differentiable in the open interval, and f(a) = f(b) given that:

[tex]f(0)=-0^2+3\,(0)=0\\f(3)=-3^2+3\,(3)=-9+9=0[/tex]

Then there must be a c in the open interval for which f'(c) =0

In order to find "c", we derive the function and evaluate it at "c", making the derivative equal zero, to solve for c:

[tex]f(x)=-x^2+3\,x\\f'(x)=-2\,x+3\\f'(c)=-2\,c+3\\0=-2\,c+3\\2\,c=3\\c=\frac{3}{2} =1.5[/tex]

There is a unique answer for c, and that is c = 1.5

Answer 2

Rolle's theorem is applicable if [tex]f(a)=f(b)[/tex] and $f$ is differentiable in $(a,b)$

since it's polynomial function, it's always continuous and differentiable..

and you can easily check that $f(0)=f(-3)=0$

so it is applicable.

now, $f'(x)=-2x+3=0 \implies x=\frac32$

there is only once value (as you can imagine, the graph will be downward parabola)

Related Questions

find the greatest common factor of 108d^2 and 216d

Answers

Answer:

Below

Step-by-step explanation:

If d is a positive number then the greatest common factor is 108d.

To get it isolate d and d^2 from the numbers.

108 divides 216. (216 = 2×108)

Then the greatest common factor of 216 and 108 is 108.

For d^2 and d we will follow the same strategy

d divides d^2 (d^2 = d*d)

Then the greatest common factor of them is d.

So the greatest common factor will be 108d if and only if d is positive. If not then 108 is the answer

Answer:

[tex]\boxed{108d}[/tex]

Step-by-step explanation:

Part 1: Find GCF of variables

The equation gives d ² and d as variables. The GCF rules for variables are:

The variables must have the same base.If one variable is raised to a power and the other is not, the GCF is the variable that does not have a power.If one variable is raised to a power and the other is raised to a power of lesser value, the GCF is the variable with the lesser value power.

The GCF for the variables is d.

Part 2: Find GCF of bases (Method #1)

The equation gives 108 and 216 as coefficients. To check for a GCF, use prime factorization trees to find common factors in between the values.

Key: If a number is in bold, it is marked this way because it cannot be divided further AND is a prime number!

Prime Factorization of 108

108 ⇒ 54 & 2

54 ⇒ 27 & 2

27 ⇒ 9 & 3

9 ⇒ 3 & 3

Therefore, the prime factorization of 108 is 2 * 2 * 3 * 3 * 3, or simplified as 2² * 3³.

Prime Factorization of 216

216 ⇒ 108 & 2

108 ⇒ 54 & 2

54 ⇒ 27 & 2

27 ⇒ 9 & 3

9 ⇒ 3 & 3

Therefore, the prime factorization of 216 is 2 * 2 * 2 * 3 * 3 * 3, or simplified as 2³ * 3³.

After completing the prime factorization trees, check for the common factors in between the two values.

The prime factorization of 216 is 2³ * 3³ and the prime factorization of 108 is 2² * 3³. Follow the same rules for GCFs of variables listed above and declare that the common factor is the factor of 108.

Therefore, the greatest common factor (combining both the coefficient and the variable) is [tex]\boxed{108d}[/tex].

Part 3: Find GCF of bases (Method #2)

This is the quicker method of the two. Simply divide the two coefficients and see if the result is 2. If so, the lesser number is immediately the coefficient.

[tex]\frac{216}{108}=2[/tex]

Therefore, the coefficient of the GCF will be 108.

Then, follow the process described for variables to determine that the GCF of the variables is d.

Therefore, the GCF is [tex]\boxed{108d}[/tex].

Let f(x) and g(x) be polynomials as shown below. Which of the following is true about f(x) and g(x)? f(x) and g(x) are not closed under multiplication because when multiplied, the result will not be a polynomial. f(x) and g(x) are not closed under multiplication because when multiplied, the result will be a polynomial. f(x) and g(x) are closed under multiplication because when multiplied, the result will not be a polynomial. f(x) and g(x) are closed under multiplication because when multiplied, the result will be a polynomial. Reset Submit

Answers

Answer:

f(x) and g(x) are closed under multiplication because when multiplied, the result will be a polynomial.

Step-by-step explanation:

The set of all polynomials is closed under addition, subtraction, and multiplication, because performing any of these operations on a pair of polynomials will give a polynomial result.

Comment on the question

The wording is a bit strange, because f(x) and g(x) are elements of a set (of polynomials), so cannot be said to be "closed." "Closed" is a property of a set with respect to some function, it is not a property of an element of the set.

Answer:

f(x) and g(x) are closed under multiplication because when multiplied, the result will be a polynomial.

Step-by-step explanation:

Its correct trust

Given the following three points, find by the hand the quadratic function they represent (0,6, (2,16, (3,33)

Answers

Answer:

[tex] f(x) = 4x^2 - 3x + 6 [/tex]

Step-by-step explanation:

Quadratic function is given as [tex] f(x) = ax^2 + bx + c [/tex]

Let's find a, b and c:

Substituting (0, 6):

[tex] 6 = a(0)^2 + b(0) + c [/tex]

[tex] 6 = 0 + 0 + c [/tex]

[tex] c = 6 [/tex]

Now that we know the value of c, let's derive 2 system of equations we would use to solve for a and b simultaneously as follows.

Substituting (2, 16), and c = 6

[tex] f(x) = ax^2 + bx + c [/tex]

[tex] 16 = a(2)^2 + b(2) + 6 [/tex]

[tex] 16 = 4a + 2b + 6 [/tex]

[tex] 16 - 6 = 4a + 2b + 6 - 6 [/tex]

[tex] 10 = 4a + 2b [/tex]

[tex] 10 = 2(2a + b) [/tex]

[tex] \frac{10}{2} = \frac{2(2a + b)}{2} [/tex]

[tex] 5 = 2a + b [/tex]

[tex] 2a + b = 5 [/tex] => (Equation 1)

Substituting (3, 33), and c = 6

[tex] f(x) = ax^2 + bx + x [/tex]

[tex] 33 = a(3)^2 + b(3) + 6 [/tex]

[tex] 33 = 9a + 3b + 6 [/tex]

[tex] 33 - 6 = 9a + 3b + 6 - 6 [/tex]

[tex] 27 = 9a + 3b [/tex]

[tex] 27 = 3(3a + b) [/tex]

[tex] \frac{27}{3} = \frac{3(3a + b)}{3} [/tex]

[tex] 9 = 3a + b [/tex]

[tex] 3a + b = 9 [/tex] => (Equation 2)

Subtract equation 1 from equation 2 to solve simultaneously for a and b.

[tex] 3a + b = 9 [/tex]

[tex] 2a + b = 5 [/tex]

[tex] a = 4 [/tex]

Replace a with 4 in equation 2.

[tex] 2a + b = 5 [/tex]

[tex] 2(4) + b = 5 [/tex]

[tex] 8 + b = 5 [/tex]

[tex] 8 + b - 8 = 5 - 8 [/tex]

[tex] b = -3 [/tex]

The quadratic function that represents the given 3 points would be as follows:

[tex] f(x) = ax^2 + bx + c [/tex]

[tex] f(x) = (4)x^2 + (-3)x + 6 [/tex]

[tex] f(x) = 4x^2 - 3x + 6 [/tex]

Why is 12 * 10-8 is NOT a correct representation of scientific notation?

Answers

Answer:

see below

Step-by-step explanation:

Scientific notation is a * 10 ^b

a must be a number between 1 ( including 1 ) and less than 10

12 is greater than 10 so it is not scientific notation

Researchers have collected data on the hours of television watched in a day and the age of a person. You are given the data below.

HOURS OF TV AGE
1 45
3 30
4 22
3 25
6 15

a. Determine which variable is the dependent variable.
b. Compute the least squares estimated line.
c. Compute the coefficient of determination. How would you interpret this value?