Simplifying Difference Quotients with Variable Expressions - How to Solve?

In summary: As for the difference quotient, it is the slope of the secant line connecting the points (a, f(a)) and (a+h, f(a+h)). So you need to write that slope formula using f(a) and f(a+h) and simplify.In summary, to find f(a), f(a + h), and the difference quotient f(a + h) − f(a), h where h ≠ 0, substitute the given values into the function f(x) = 5x/(x-1). For f(a) and f(a + h), simply replace x with the given values, and for the difference quotient, use the slope formula connecting the points (a, f(a)) and (a+h,
  • #1
Niaboc67
249
3

Homework Statement


Find f(a), f(a + h),and the difference quotient f(a + h) − f(a), h where h ≠ 0.

f(x) = 5x/x-1

f(a) = ?
f(a + h) = ?
f(a + h) − f(a)/(h) = ?


The Attempt at a Solution



As I understand it I must substitute the values but I don't understand the question. say if it were something like 1/x I would know to fill it into (1/(x+h)-1/x) / h
But this just confuses me I don't know how to answer these questions
 
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  • #2
Niaboc67 said:

Homework Statement


Find f(a), f(a + h),and the difference quotient f(a + h) − f(a), h where h ≠ 0.

f(x) = 5x/x-1

f(a) = ?
f(a + h) = ?
f(a + h) − f(a)/(h) = ?


The Attempt at a Solution



As I understand it I must substitute the values but I don't understand the question. say if it were something like 1/x I would know to fill it into (1/(x+h)-1/x) / h
But this just confuses me I don't know how to answer these questions

The way you wrote it you have
[tex] f(x) = \frac{5x}{x} - 1 = 5 - 1 = 4 [/tex]
so ##f(a) = f(a+h) = 4##. Perhaps you really did not mean what you wrote, but because you failed to use parentheses it is impossible to say. And, of course, you wrote
[tex]f(a+h) - \frac{f(a)}{h} = 4 - \frac{4}{h}. [/tex]
Again, maybe you meant something different.
 
  • #3
What I meant was: f(x) = (5x)/(x-1) my mistake
 
  • #4
Niaboc67 said:
What I meant was: f(x) = (5x)/(x-1) my mistake
Just replace every instance of x with the given value. Where the given value consists of an expression, like a+h, put the expression in parentheses when making the substitution.
 

What are difference quotient problems?

Difference quotient problems are mathematical problems that involve finding the average rate of change of a function over a specific interval. They are also known as "slope problems" or "rate of change problems."

How do I solve difference quotient problems?

To solve a difference quotient problem, you first need to find the slope of the function at a given point. This can be done by finding the difference in y-values over the difference in x-values for two points on the function. Then, you can plug in the values into the difference quotient formula: (f(x+h) - f(x)) / h. Simplify the equation to solve for the average rate of change.

What is the purpose of difference quotient problems?

The purpose of difference quotient problems is to calculate the average rate of change of a function over a specific interval. This can be useful in real-life situations, such as determining the average speed of an object or the average growth rate of a population.

Are there any common mistakes when solving difference quotient problems?

Yes, there are some common mistakes that people make when solving difference quotient problems. These include mixing up the order of the points, forgetting to simplify the equation, and not using the correct formula. It is important to double check your work to ensure accuracy.

Can difference quotient problems be used for any type of function?

Yes, difference quotient problems can be used for any type of function, including linear, quadratic, exponential, and trigonometric functions. The only requirement is that the function must be continuous over the given interval.

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