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Help with this derivative?
The derivative of the function F(x)= x/(x^2 +2x- 3) is (-x^2 - 3)/(x^2 +2x -3)^2
Now, I want to solve F'(x)=0 to complete the table of variations and deduce the sense of variation and plot the graph. When I solve F'(x)=0 it turns out as an impossible answer of x^2 = -3... What can I do in such case?
3 Answers
- 8 years agoFavorite Answer
Hi,
The differentiation of the function is absolutely correct.
Now, if you're trying to equate F'(x)=0 it means you're trying to find the intervals or the values for which F(x) is increasing or decreasing.
Putting F'(x)=0, you get:
-x^2=3
x^2=-3
Now, since the the square root of a negative number is undefined, the answer to this question is as follows: for no value of x belonging to real number (R) is F(x) increasing or decreasing. To put it simply, the graph of F(x) will neither be increasing, nor decreasing.
But, if you're aware of numbers called 'complex' numbers, then your answer would be:
x^2=-3
x=3i where i=(-1)^1/2
And on the Argand plane, the function F(x) can be expressed as it accommodates complex values as well as real values of x.
Check Source(s) for a link to the Wiki on Complex Numbers.
Hope this helps! :)
(5 stars?)
Source(s): http://en.wikipedia.org/wiki/Complex_number - SS4Lv 78 years ago
It means the F(x) is not differentiable at 0. The graph of the derivative supports this
- 8 years ago
It is not impossible. Remember the imaginary number i is defined as i^2=-1
That means x=3i
