Calculating Instantaneous Rate of Change for a Quadratic Function

[Jen23]

Homework Statement

Estimate the instantaneous rate of change of the function f(x)=3x^2 + 4x at (1,7)

Homework Equations

∆f(x)/∆x = f(x2)-f(x1) / x2-x1

The Attempt at a Solution

I know that x=1 given the point, but to find the instantaneous rate of change I can use x=1.001 as this is a very close number to the point on the tangent curve. So:
∆f(x)/∆x = [3(1.001)^2 + 4(1.001)] - [3(1)^2 + 4(1) ] / 1.001 - 1
= 7.010003 - 7 / 1.001 - 1
= 10.003

The answer in the book is 13 for instantaneous rate of change, and they used the interval 1 ≤ x ≤ 2
I do not understand why they chose to use the point 2, I thought that the point on a tangent must be very close to the x value so wouldn't 1.001 be more appropriate? Have I answered the question correctly? Detailed explanation would be appreciated.

 

[magoo]

Your methodology is OK, but it doesn't lend itself to an estimate. It is a detailed precise calculation.

If you plot the function with x varying from o to 3 in steps of 1, I think you can see why the author chose the evaluation at x = 1 and then x = 2. You can quickly figure this in your head.

So your answer is not wrong, it simply is not an estimate. It is more precise that what the author came up with, but it is not an estimate because you evaluated the function at x = 1.001 - which you really need a calculator or worksheet to do.

 

[Jen23]

Your methodology is OK, but it doesn't lend itself to an estimate. It is a detailed precise calculation.

If you plot the function with x varying from o to 3 in steps of 1, I think you can see why the author chose the evaluation at x = 1 and then x = 2. You can quickly figure this in your head.

So your answer is not wrong, it simply is not an estimate. It is more precise that what the author came up with, but it is not an estimate because you evaluated the function at x = 1.001 - which you really need a calculator or worksheet to do.

Okay I see what you are saying, thanks for taking the time to read and reply to my question!

 

[Mark44]

Homework Statement

Estimate the instantaneous rate of change of the function f(x)=3x^2 + 4x at (1,7)

Homework Equations

∆f(x)/∆x = f(x2)-f(x1) / x2-x1

You really need parentheses here. What you wrote on the right side means $f(x2) - \frac{f(x1)}{x2} - x1$. I'm sure that's not what you meant.

The Attempt at a Solution

I know that x=1 given the point, but to find the instantaneous rate of change I can use x=1.001 as this is a very close number to the point on the tangent curve.

But, that's not very close to 1.7, which is what I think you meant by writing "at (1,7)".

So:
∆f(x)/∆x = [3(1.001)^2 + 4(1.001)] - [3(1)^2 + 4(1) ] / 1.001 - 1
= 7.010003 - 7 / 1.001 - 1
= 10.003

The answer in the book is 13 for instantaneous rate of change, and they used the interval 1 ≤ x ≤ 2
I do not understand why they chose to use the point 2, I thought that the point on a tangent must be very close to the x value so wouldn't 1.001 be more appropriate? Have I answered the question correctly? Detailed explanation would be appreciated.

As best as I can determine they want an estimate at x = 1.7, not around x = 1.

 

[Mark44]

Your methodology is OK, but it doesn't lend itself to an estimate. It is a detailed precise calculation.

Precise or not, it's still an estimate. The OP is estimating the slope of the tangent line by the slope of a secant line.

If you plot the function with x varying from o to 3 in steps of 1, I think you can see why the author chose the evaluation at x = 1 and then x = 2. You can quickly figure this in your head.

So your answer is not wrong, it simply is not an estimate.

No, it is an estimate of the slope of the tangent line (i.e., the instantaneous rate of change at a particular point).

It is more precise that what the author came up with, but it is not an estimate because you evaluated the function at x = 1.001 - which you really need a calculator or worksheet to do.

The OP evaluated the function in order to estimate the value of its derivative.
A calculator is nice, but not necessary. One can square 1.001 using only pencil and paper.

 

[Jen23]

Precise or not, it's still an estimate. The OP is estimating the slope of the tangent line by the slope of a secant line.
No, it is an estimate of the slope of the tangent line (i.e., the instantaneous rate of change at a particular point).

The OP evaluated the function in order to estimate the value of its derivative.
A calculator is nice, but not necessary. One can square 1.001 using only pencil and paper.

Yes, that is what I was thinking, I had just calculated and estimate the slope of the tangent line. We can calculate this instantaneous rate as x approaches that certain value (example, using 1.0001, 1.00001,..). I included four decimal places at least for accuracy.

The book did use the point (1, 7) I think after inserting x=1 into the function we get 7, so f(1)=7. I chose a point x=1.001 (point on the curve that is very close to 1) and solved for f(1.001)= 7.010003. After finding the slope of these two points, it gives me an estimate of the instantaneous rate of change, I assume?

 

[Mark44]

I don't know why they used 1 and 2 in their approximation. To get a better approximation, you would normally pick values on either side of 1, such as 0 and 2 or .5 and 1.5. As both numbers get closer to 1, the approximation gets better.

The exact value of the inst. rate of change (the derivative, in other words) at x = 1 is 10, so your approximations are pretty close. Their approximation of 13 is pretty far off.

 

[Jen23]

I don't know why they used 1 and 2 in their approximation. To get a better approximation, you would normally pick values on either side of 1, such as 0 and 2 or .5 and 1.5. As both numbers get closer to 1, the approximation gets better.

The exact value of the inst. rate of change (the derivative, in other words) at x = 1 is 10, so your approximations are pretty close. Their approximation of 13 is pretty far off.

I also thought using 1 and 2 is on the farther end of getting a better approximation. I am going to stick to what I originally came up with since it is a closer approximation, maybe the book had an error. Thank you for your feedback! I appreciate it.

 

[Chestermiller]

For a quadratic function, if you choose two points equidistant on either side of the desired point (x = 1.7), the finite divided difference will give you the exact value of the rate of change at the desired point. For example, if you choose x1 = 0 and x2 = 3.4, the finite divided difference gives you a rate of change of 14.2, which is the exact value. Try x1 = 0 and x2 = 2.4 and see what you get.