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3 years ago

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Function f(x) is positive, increasing and concave up on the closed interval [a, b]. The interval [a, b] is partitioned into 4 eq

ual intervals and these are used to compute the upper sum, lower sum, and trapezoidal rule approximations for the value of Integral b a f(x) dx. Which one of the following statements is true?
Lower sum < Trapezoidal rule Value < Upper sum
Lower sum < Upper sum < Trapezoidal rule value
Trapezoidal rule < Lower sum < Upper sum
Cannot be determined without the x-values for the partitions

2 answers:

Umnica [9.8K]3 years ago

5 0

The left sum would be f0+f1+f2+f3

The right sum would be f1+f2+f3+f4

The trapezoidal rule value is:

(f0+f1)/2 + (f1+f2)/2+(f2+f3)/2 +(f3+f4)/2

This would put the trapezoidal rule in the middle , which makes the answer:

Lower sum < Trapezoidal rule Value < Upper sum

timama [110]3 years ago

5 0

Answer:

Step-by-step explanation:

Given function f(x) is positive, increasing and concave up on the closed interval [a, b],

it means f(x1) < f(x2) if x1 < x2

So Lower sum < Upper sum

As Trapezoidal is average of f(x1) and f(x2) = [f(x1) + f(x2)] / 2

it is average of the Lower and Upper sum.

The answer is Lower sum < Trapezoidal rule Value < Upper sum

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Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x3 − 6x2 − 15x + 2 (a) Find the interval on which

Burka [1]

Answer:

a) $(-\infty, -1) \cup (5, \infty)$

b) $(-1,5)$

Step-by-step explanation:

The first step to solve this question is finding the roots of the derivative of x.

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

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$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

$f(x) = x^{3} - 6x^{2} - 15x + 2$

So

$f'(x) = 3x^{2} - 12x - 15$

Finding the roots:

$3x^{2} - 12x - 15 = 0$

Simplifying by -3

$x^{2} - 4x - 5 = 0$

So $a = 1, b = -4, c = -5$

Then

$\bigtriangleup = (-4)^{2} - 4*1*(-5) = 36$

$x_{1} = \frac{-(-4) + \sqrt{36}}{2} = 5$

$x_{2} = \frac{-(-4) - \sqrt{36}}{2} = -1$

So the function can be divided in three intervals.

They are:

Less than -1

Between -1 and 5

Higher than 5

In which it increases and which it decreases?

Less than -1

Lets find the derivative in a point in this interval, for example, -2

$f'(x) = 3x^{2} - 12x - 15$

$f'(-2) = 3*(-2)^{2} - 12*(-2) - 15 = 21$

Positive.

So in the interval of $(-\infty, -1)$ , the function increases.

Between -1 and 5

Will choose 0.

$f'(x) = 3x^{2} - 12x - 15$

$f'(0) = 3*(0)^{2} - 12*(0) - 15 = -15$

Negative.

So in the interval of $(-1,5)$ , the function decreases.

Higher than 5

Will choose 6.

$f'(x) = 3x^{2} - 12x - 15$

$f'(6) = 3*(6)^{2} - 12*(6) - 15 = 21$

Positive

So in the interval of $(5, \infty)$ , the function increases.

(a) Find the interval on which f is increasing.

Using interval notation

$(-\infty, -1) \cup (5, \infty)$

b) Find the interval on which f is decreasing.

$(-1,5)$

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