# Calculating Instantaneous Rate of Change for a Quadratic Function

• Jen23
In summary: The OP wrote "the instantaneous rate of change" and "this is a very close number to the point on the tangent curve".In summary, the conversation discussed estimating the instantaneous rate of change of the function f(x)=3x^2 + 4x at (1,7). The method used involved evaluating the function at a very close number to the point on the tangent curve, x=1.001. However, it was noted that this was not a good estimate
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.

Last edited:
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.

magoo said:
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!

Jen23 said:

## 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.
Jen23 said:

## 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)".
Jen23 said:
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.

magoo said:
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.
magoo said:
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).
magoo said:
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.

Mark44 said:
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?

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.

Mark44 said:
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.

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.

SammyS

## What is the instantaneous rate of change?

The instantaneous rate of change is the rate at which a function changes at a specific point. It represents the slope of the tangent line at that point on the function's graph.

## How is the instantaneous rate of change different from average rate of change?

The average rate of change is the overall change in a function over a given interval, while the instantaneous rate of change is the change at a specific point. The average rate of change is represented by the slope of a secant line, while the instantaneous rate of change is represented by the slope of the tangent line at a point.

## What is the formula for calculating instantaneous rate of change?

The formula for calculating instantaneous rate of change is the derivative of the function at a specific point. It can be written as f'(x) or dy/dx, where x is the point at which the instantaneous rate of change is being calculated.

## What is the significance of instantaneous rate of change in real-world applications?

The instantaneous rate of change is used in many real-world applications, such as calculating the velocity of an object at a specific point in time or the rate of change of a chemical reaction at a certain temperature. It helps us understand the behavior of functions and how they change over time or in different conditions.

## How do you interpret the value of instantaneous rate of change?

The value of instantaneous rate of change tells us the exact rate of change at a specific point on a function. A positive value indicates that the function is increasing at that point, while a negative value indicates that it is decreasing. A value of 0 indicates a point of inflection, where the function changes from increasing to decreasing or vice versa.

Replies
3
Views
1K
Replies
23
Views
3K
Replies
1
Views
2K
Replies
3
Views
7K
Replies
4
Views
2K
Replies
30
Views
2K
Replies
2
Views
1K
Replies
2
Views
789
Replies
2
Views
2K
Replies
4
Views
958