We are interested in when the second derivative is equal to zero. This actually happens a few times with our data set, but there is only one equivalence point. With our data there are a number of consecutive places early on where the second derivative is zero. To figure out why this happens, suppose we have a function whose second derivative is always zero. So we have dy 0. What is for this function? fip dx dy dr? With that derivative, what kind of function must y be? Conclusion: If f(x) is a function on an interval, then f"(x) = on that interval.

Calculus: Early Transcendentals
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Author:James Stewart
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Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Left Column is mL of NaOH

Right Column is pH level

We are interested in when the second derivative is equal to zero. This actually happens a few times with our data set, but there is only one equivalence point. With our data there are a number of consecutive places early on where the second derivative is zero. To figure out why this happens, suppose we have a function whose second derivative is always zero. So we have

\[ \frac{d^2y}{dx^2} = 0. \] 

What is \( \frac{dy}{dx} \) for this function?

With that derivative, what kind of function must \( y \) be? Conclusion: If \( f(x) \) is a _________________ function on an interval, then \( f''(x) = \,  \_\_\_\_ \) on that interval.
Transcribed Image Text:We are interested in when the second derivative is equal to zero. This actually happens a few times with our data set, but there is only one equivalence point. With our data there are a number of consecutive places early on where the second derivative is zero. To figure out why this happens, suppose we have a function whose second derivative is always zero. So we have \[ \frac{d^2y}{dx^2} = 0. \] What is \( \frac{dy}{dx} \) for this function? With that derivative, what kind of function must \( y \) be? Conclusion: If \( f(x) \) is a _________________ function on an interval, then \( f''(x) = \, \_\_\_\_ \) on that interval.
The table below shows the relationship between the volume of NaOH (in mL) added and the pH of the solution. This data is typically used to understand the titration process between a strong base (NaOH) and a weak acid. 

| mL of NaOH | pH   |
|------------|------|
| 0          | 2.22 |
| 5          | 2.83 |
| 8          | 3.09 |
| 10         | 3.23 |
| 13         | 3.41 |
| 15         | 3.53 |
| 18         | 3.71 |
| 20         | 3.83 |
| 25         | 4.24 |
| 27         | 4.51 |
| 28         | 4.72 |
| 29         | 5.10 |
| 30         | 10.66|
| 31         | 11.37|
| 33         | 11.77|
| 35         | 11.96|
| 40         | 12.22|

At the beginning of the titration process (0 mL of NaOH), the pH is 2.22, indicating an acidic solution. As NaOH is added, the pH gradually increases, showing the neutralization of the acid. A dramatic increase in pH is observed between the addition of 29 mL and 30 mL of NaOH, which is typical at the equivalence point in a titration. Beyond this point, the pH levels off as the solution becomes more basic with continued NaOH addition.
Transcribed Image Text:The table below shows the relationship between the volume of NaOH (in mL) added and the pH of the solution. This data is typically used to understand the titration process between a strong base (NaOH) and a weak acid. | mL of NaOH | pH | |------------|------| | 0 | 2.22 | | 5 | 2.83 | | 8 | 3.09 | | 10 | 3.23 | | 13 | 3.41 | | 15 | 3.53 | | 18 | 3.71 | | 20 | 3.83 | | 25 | 4.24 | | 27 | 4.51 | | 28 | 4.72 | | 29 | 5.10 | | 30 | 10.66| | 31 | 11.37| | 33 | 11.77| | 35 | 11.96| | 40 | 12.22| At the beginning of the titration process (0 mL of NaOH), the pH is 2.22, indicating an acidic solution. As NaOH is added, the pH gradually increases, showing the neutralization of the acid. A dramatic increase in pH is observed between the addition of 29 mL and 30 mL of NaOH, which is typical at the equivalence point in a titration. Beyond this point, the pH levels off as the solution becomes more basic with continued NaOH addition.
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