Difference Quotient Isn't working.

  • #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?
 

Answers and Replies

  • #2
Gib Z
Homework Helper
3,352
6
[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
Dickfore
2,988
5
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
themadhatter1
140
0
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
Dickfore
2,988
5
No. Please review limits before you go on to derivatives.
 
  • #6
themadhatter1
140
0
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
HallsofIvy
Science Advisor
Homework Helper
43,021
970
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
Dickfore
2,988
5
I think you mean "indeterminant" form.

I think you meant "indeterminate" form.
 
  • #10
Mentallic
Homework Helper
3,802
94
Hahaha gold :rofl:
 

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