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1 year ago

H(x)=

Mathematics

1 answer:

UNO [17]1 year ago

7 0

The average rate of change of the function over −2 ≤ x ≤ 2 is 1/2

<h3>How to determine the average rate of change?</h3>

The function is given as:

$h(x) = \frac 18x^3 - x^2$

The interval is given as:

−2 ≤ x ≤ 2

Calculate h(2) and h(-2)

$h(2) = \frac 18 * 2^3 - (2)^2$

h(2) = -3

$h(-2) = \frac 18 * (-2)^3 - (-2)^2$

h(-2) = -5

The average rate of change is then calculated as:

$m = \frac{h(-2) - h(2)}{-2-2}$

This gives

$m = \frac{-5 + 3}{-4}$

Evaluate

m = 1/2

Hence, the average rate of change of the function over −2 ≤ x ≤ 2 is 1/2

Read more about average rate of change at

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<u>Complete question</u>

$h(x) = \frac 18x^3 - x^2$

What is the average rate of change of h over the interval −2 ≤ x ≤ 2

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The base of the triangle decreases at a rate of 2.262 centimeters per minute.

Step-by-step explanation:

From Geometry we understand that area of triangle is determined by the following expression:

$A = \frac{1}{2}\cdot b\cdot h$ (Eq. 1)

Where:

$A$ - Area of the triangle, measured in square centimeters.

$b$ - Base of the triangle, measured in centimeters.

$h$ - Height of the triangle, measured in centimeters.

By Differential Calculus we deduce an expression for the rate of change of the area in time:

$\frac{dA}{dt} = \frac{1}{2}\cdot \frac{db}{dt}\cdot h + \frac{1}{2}\cdot b \cdot \frac{dh}{dt}$ (Eq. 2)

Where:

$\frac{dA}{dt}$ - Rate of change of area in time, measured in square centimeters per minute.

$\frac{db}{dt}$ - Rate of change of base in time, measured in centimeters per minute.

$\frac{dh}{dt}$ - Rate of change of height in time, measured in centimeters per minute.

Now we clear the rate of change of base in time within (Eq, 2):

$\frac{1}{2}\cdot\frac{db}{dt}\cdot h = \frac{dA}{dt}-\frac{1}{2}\cdot b\cdot \frac{dh}{dt}$

$\frac{db}{dt} = \frac{2}{h}\cdot \frac{dA}{dt} -\frac{b}{h}\cdot \frac{dh}{dt}$ (Eq. 3)

The base of the triangle can be found clearing respective variable within (Eq. 1):

$b = \frac{2\cdot A}{h}$

If we know that $A = 130\,cm^{2}$ , $h = 15\,cm$ , $\frac{dh}{dt} = 2.5\,\frac{cm}{min}$ and $\frac{dA}{dt} = 4.7\,\frac{cm^{2}}{min}$ , the rate of change of the base of the triangle in time is:

$b = \frac{2\cdot (130\,cm^{2})}{15\,cm}$

$b = 17.333\,cm$

$\frac{db}{dt} = \left(\frac{2}{15\,cm}\right)\cdot \left(4.7\,\frac{cm^{2}}{min} \right) -\left(\frac{17.333\,cm}{15\,cm} \right)\cdot \left(2.5\,\frac{cm}{min} \right)$

$\frac{db}{dt} = -2.262\,\frac{cm}{min}$

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

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