Difference Quotient Isn't working.

In summary, the conversation is about finding the derivative of a given function using the difference quotient and using that derivative to determine any points on the graph where the tangent line is horizontal. The conversation also involves discussing the rules of differentiation and clarifying any errors in the process.
  • #1
themadhatter1
140
0

Homework Statement


Find the derivative of f using the difference quotient and use the derivative of f to determine any points on the graph of f where the tangent line is horizontal.

[tex]f(x)=3x^3-9x[/tex]

Homework Equations




The Attempt at a Solution



[tex]\lim_{\Delta x\rightarrow0}=\frac{3(x+\Delta x)^3-9(x+\Delta x)-3x^3+9x}{\Delta x}[/tex]


[tex]\lim_{\Delta x\rightarrow0}=\frac{3(x^3+3x^2\Delta x+3x\Delta x^2+\Delta x^3)-9x-9\Delta x-3x^3+9x}{\Delta x}[/tex]


[tex]\lim_{\Delta x\rightarrow0}=\frac{3x^3+9x^2\Delta x+9x\Delta x^2+3\Delta x^3-9x-9\Delta x-3x^3+9x}{\Delta x}[/tex]

The '3x3's and '9x's cancel out and you are left with


[tex]\lim_{\Delta x\rightarrow0}=\frac{9x^2\Delta x+9x\Delta x^2+3\Delta x^3-9\Delta x}{\Delta x}[/tex]

Then if you take the limit as delta x approaches 0 you will get

[tex]9x^2+9x-6[/tex] I know the answer is [tex]9x^2-9[/tex] via the rules of differentiation.

What am I doing wrong with this example?
 
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  • #2
themadhatter1 said:
[tex]\lim_{\Delta x\rightarrow0}=\frac{9x^2\Delta x+9x\Delta x^2+3\Delta x^3-9\Delta x}{\Delta x}[/tex]

Then if you take the limit as delta x approaches 0 you will get

[tex]9x^2+9x-6[/tex] I know the answer is [tex]9x^2-9[/tex] via the rules of differentiation.

What am I doing wrong with this example?

Check the bolded step again.
 
  • #3
You can cancel factors of [itex]\Delta x[/itex] from each term in the numerator with the one in the denominator. What are you left with after this cancellation? What is the limit of this expression when [itex]\Delta x \rightarrow 0[/itex]?
 
  • #4
Dickfore said:
You can cancel factors of [itex]\Delta x[/itex] from each term in the numerator with the one in the denominator. What are you left with after this cancellation? What is the limit of this expression when [itex]\Delta x \rightarrow 0[/itex]?

[tex]\lim_{\Delta x\rightarrow0}=\frac{9x^2\Delta x+9x\Delta x^2+3\Delta x^3-9\Delta x}{\Delta x}[/tex]

Ok, so you would get this after taking the limit.

[tex]=9x^2+9x\Delta x+3\Delta x^2-9[/tex]

what happens to the delta x terms though? Is delta x just such a small number [tex]\epsilon[/tex] that you can say they are practically 0 and anything multiplied by 0 is 0?
 
  • #5
No. Please review limits before you go on to derivatives.
 
  • #6
Nevermind you would simply factor out a delta x out of the numerator then you have 2 'delta x's that cancel out. Then you take the limit and you get 9x2-9. Otherwise its in indiscriminate form.
 
  • #7
themadhatter1 said:
Nevermind you would simply factor out a delta x out of the numerator then you have 2 'delta x's that cancel out. Then you take the limit and you get 9x2-9. Otherwise its in indiscriminate form.
I think you mean "indeterminant" form.
 
  • #8
HallsofIvy said:
I think you mean "indeterminant" form.

I think you meant "indeterminate" form.
 
  • #9
HallsofIvy said:
I think you mean "indeterminant" form.

Dickfore said:
I think you meant "indeterminate" form.
HallsOfIvy was being indiscriminate.:smile:
 
  • #10
Hahaha gold :rofl:
 

1. Why is the Difference Quotient not giving accurate results?

The Difference Quotient may not be giving accurate results because the function you are using is not continuous or differentiable at the specific point you are evaluating. This means that the formula for the Difference Quotient, which involves taking the limit as the change in x approaches 0, may not exist or may not accurately represent the function's behavior at that point.

2. How can I fix the issue with the Difference Quotient?

If you are encountering issues with the Difference Quotient, you can try using a different method for finding the derivative such as the Power Rule, Product Rule, or Quotient Rule. You can also try using a smaller interval for the change in x when evaluating the Difference Quotient, or check if the function is continuous and differentiable at the point you are evaluating.

3. Can the Difference Quotient be used for all types of functions?

No, the Difference Quotient is only applicable for continuous and differentiable functions. If a function is not continuous or differentiable at a specific point, the Difference Quotient cannot be used to find the derivative at that point.

4. What are some common mistakes when using the Difference Quotient?

Some common mistakes when using the Difference Quotient include using an incorrect formula, not simplifying the expression before taking the limit, or using an interval for the change in x that is too large. It is important to carefully follow the steps of the Difference Quotient and check for any errors in your calculations.

5. Are there any alternative methods to the Difference Quotient?

Yes, there are several alternative methods for finding the derivative of a function, such as the Power Rule, Product Rule, and Quotient Rule. These methods may be more applicable and accurate for certain types of functions, so it is important to understand and use them in addition to the Difference Quotient.

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