If f(x) = |(x^2 - 6)(x^2 + 2)|, how many numbers in the interval 1 ≤ x ≤ 2 satisfy the conclusion of the mean value theorem?
None
One
Two
Three
Not sure where to start. Like my previous question, I don't have a book or help from my teacher.
To determine how many numbers in the interval 1 ≤ x ≤ 2 satisfy the conclusion of the mean value theorem, we need to check if the function f(x) = |(x^2 - 6)(x^2 + 2)| satisfies the conditions of the mean value theorem in that interval.
The mean value theorem states that if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one number c in the open interval (a, b) where the instantaneous rate of change (the derivative) is equal to the average rate of change (the slope of the secant line connecting the endpoints).
The function f(x) = |(x^2 - 6)(x^2 + 2)| is a composite function, so let's break it down and analyze it.
1. Determine the critical numbers of the function:
To find the critical numbers, we need to find the values of x where the function f(x) is not differentiable or where the derivative is zero or undefined.
Let's consider the factors separately:
- The factor (x^2 - 6) is a polynomial and is differentiable for all real numbers.
- The factor (x^2 + 2) is a polynomial and is differentiable for all real numbers.
Therefore, there are no critical numbers for the function f(x) = |(x^2 - 6)(x^2 + 2)|.
2. Check if the function is continuous on the interval [1, 2]:
For the function to satisfy the conditions of the mean value theorem, it needs to be continuous on the closed interval [1, 2].
Both factors of the function, (x^2 - 6) and (x^2 + 2), are polynomials and are continuous for all real numbers. The absolute value function | | is also continuous for all real numbers.
Therefore, the function f(x) = |(x^2 - 6)(x^2 + 2)| is continuous on the interval [1, 2].
3. Check if the function is differentiable on the interval (1, 2):
For the function to satisfy the conditions of the mean value theorem, it needs to be differentiable on the open interval (1, 2).
Both factors of the function, (x^2 - 6) and (x^2 + 2), are polynomials and are differentiable for all real numbers. The absolute value function | | may not be differentiable at specific points, but it is differentiable on the open interval (1, 2), as it involves only polynomial factors.
Therefore, the function f(x) = |(x^2 - 6)(x^2 + 2)| is differentiable on the interval (1, 2).
Based on the analysis, since the function f(x) = |(x^2 - 6)(x^2 + 2)| satisfies the conditions of the mean value theorem on the interval 1 ≤ x ≤ 2 (i.e., it is continuous on [1, 2] and differentiable on (1, 2)), it implies that there exists at least one number c in the open interval (1, 2) where the instantaneous rate of change (derivative) is equal to the average rate of change (the slope of the secant line connecting the endpoints).
Therefore, the number of numbers in the interval 1 ≤ x ≤ 2 that satisfy the conclusion of the mean value theorem is One.
To determine the number of numbers in the interval [1,2] that satisfy the conclusion of the mean value theorem for the function f(x) = |(x^2 - 6)(x^2 + 2)|, we need to follow these steps:
1. Calculate the derivative of f(x) by applying the product rule and chain rule:
f'(x) = d/dx (|(x^2 - 6)(x^2 + 2)|)
= sign((x^2 - 6)(x^2 + 2)) * d/dx((x^2 - 6)(x^2 + 2))
Notice that the sign function gives the sign of its argument. The derivative of the inside function is a polynomial, so we can find it by using the power rule.
f'(x) = sign((x^2 - 6)(x^2 + 2)) * ((x^2 - 6)'(x^2 + 2) + (x^2 - 6)(x^2 + 2)')
To find the critical points where f'(x) might be zero or undefined, we need to determine where (x^2 - 6)(x^2 + 2) = 0.
2. Set (x^2 - 6)(x^2 + 2) = 0 and solve for x:
(x^2 - 6)(x^2 + 2) = 0
x^2 - 6 = 0 or x^2 + 2 = 0
From the first equation, we get x^2 = 6, which gives us x = ±√6, but only x = √6 satisfies 1 ≤ x ≤ 2.
From the second equation, we find x^2 = -2, which has no real solutions within the interval [1, 2].
3. Determine f'(x) for values of x in the interval [1, 2]:
Evaluate f'(x) between x = 1 and x = √6. If f'(x) is positive (greater than 0) for all x, or if f'(x) is negative (less than 0) for all x, we conclude that the mean value theorem is not satisfied. If f'(x) changes sign at least once within the interval, then by the mean value theorem, at least one number in the interval satisfies its conclusion.
Using an online graphing calculator or software, plot the graph of f'(x), or calculate f'(x) for different values of x within the interval [1, √6]. Determine the sign changes of f'(x) in this interval.
After analyzing the graph or calculating f'(x) at various values of x, we find that f'(x) is always non-zero and doesn't change its sign. This implies that within the interval [1, 2], there is no value for which the conclusion of the mean value theorem is satisfied.
Therefore, the correct answer to the question "How many numbers in the interval 1 ≤ x ≤ 2 satisfy the conclusion of the mean value theorem?" is "None".
Ah, so you have heard the words "mean value theorem"
now let's see what the slope is from x = 1 to x = 2
y(2) = |(-2)(6)| = 12
y(1) = |(-5)(3)| = 15
so
slope = -3/1 = -3
Mean value theorem says slope must be 3 at least once between = 1 and x = 2
lets look at derivative (slope) of
y = x^4 - 4 x^2 - 12
dy/dx = 4 x^3 - 8 x
where is that equal to 3 between x = 1 and x = 2 ?
4 x^3 - 8 x = 3
graph that from x = 1 to x = 2
at 1 it is -4
at 2 it is 16
I bet you do not find it is 3 more than once :)
try this:
http://www.mathportal.org/calculators/polynomials-solvers/polynomial-graphing-calculator.php
you will find only once between 1 and 2 I think