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# I don't really know how to work with intervals so:

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I don't really know how to work with intervals so:

Given is the function f(x)=2x^3 - x^2 - 6x+12

a. Calculate to one decimal place the average change of the function over the interval [0, 12]

b. Given is that the differential quotient of f on the interval [0, a] equals 0. How do I calculate a?

c. Approximate the slope of the graph of f at the point with x-coordinate 3. Let Δx be 0.001

d. On the interval [(-1/2) (1/2] there is a point P on the graph of f where the slope of the tangent line is the smallest. Determinte the coordinates of P to one decimal place and please explain what you did.

Sep 9, 2020

#1
+110689
+1

I don't really know how to work with intervals so:

Given is the function f(x)=2x^3 - x^2 - 6x+12

a. Calculate to one decimal place the average change of the function over the interval [0, 12]

You are being asked to find    f(12)  - f(0)

And then to divide the answer by 12-0

b. Given is that the differential quotient of f on the interval [0, a] equals 0. How do I calculate a?

Same as above only this time the answer is 0

c. Approximate the slope of the graph of f at the point with x-coordinate 3. Let Δx be 0.001

Same as (a)  only using   the interval [3, 3.001]

d. On the interval [(-1/2) (1/2] there is a point P on the graph of f where the slope of the tangent line is the smallest. Determinte the coordinates of P to one decimal place and please explain what you did.

I suggest you graph your function so that you can understand what you have done.

If you do some yourself, and display it, then you can ask more guidance if you want to.

Sep 9, 2020
#2
+1

Hi thank you for the reply!

f(x)=2x^3 - x^2 - 6x+12

a.f(12)=2(12)^3 - (12)^2 - 6(12)+12 = 3252

f(0)=2(0)^3 - (0)^2 - 6(0)+12 = 12

3252-12 = 3240

3240/12 = 270

So the average change is just 270.0?

c. f(3)=2(3)^3 - (3)^2 - 6(3)+12 = 39

f(3.001)=2(3.001)^3 - (3.001)^2 - 6(3.001)+12 = 39.042017002

39 - 39.042017002 = -0.042017002

-0.042017002/-0.042017002 = 1

I'm guessing I'm doing something wrong here and misunderstanding your directions at a. ?

Guest Sep 9, 2020
#3
+110689
+1

A gradient = rise on run = difference in y values over difference in x values.

https://www.geogebra.org/classic/zmc4kuwp

Your part c is not right because you have not divided by the difference in the x values.

the x values are   3.001 and 0.001 so the difference is 0.001

Melody  Sep 9, 2020
#4
+110689
0

If the link doesn't work for you then let me know.

Melody  Sep 9, 2020
#5
+1

The link works, thank you! I'm normally not amazing at understanding the concept behind what I'm calculating but I think I do this time. So for c. it would be

-0.042017002/0.001 = -42.017002

right? This might be me being confused again but since they're asking for me to approximate the slope would that even be correct?

I'm also not completely understanding what you said at b. Is a just 0?

Guest Sep 10, 2020
#6
+110689
+1

Unfortunately I have already thrown out the working that I did for your question....

oh well..

the points are   (3, 39) and  (3.001, 39.042)               rounded more

the gradient between these points will be    $$\frac{39.042-39}{3.001-3}$$

Can you see what you have done wrong?

It is an approximation because they actually asked you for the gradient of the tangent at (3, 39)

And you have found the gradient of the secant between (3,39) and (3.001, 39.042)

You could just as easily found the gradient of the secant between the points where x=3 and x=2.999.  That would have been slightly different but also an acceptable estimation.

Part b

b. Given is that the differential quotient of f on the interval [0, a] equals 0. How do I calculate a?

This just means that the gradient of the line joining the 2 points, one where x=0 and the other point where x=a has a gradient of 0

$$f(0)=12\\ f(a)=2a^3-a^2-6a+12\\ so\\ \frac{f(a)-f(0)}{a-0}=0\\ \frac{2a^3-a^2-6a+12-12}{a-0}=0\\ \frac{2a^3-a^2-6a}{a}=0\\$$

Can you take it from there?

Melody  Sep 10, 2020