Asked by Sam
Find the point where the tangent to the curve is horizontal.
f(x) = (x^2-2x+4)/(x^2+4)
f(x) = (x^2-2x+4)/(x^2+4)
Answers
Answered by
drwls
That would be there the derivative
f'(x) = 0
The derivative is
f'(x) = (x^2+4)(2x -2) - (x^2-2x+4)(2x)]/(x^2 +4)^2
Set the numerator = 0 and solve for x.
f'(x) = 0
The derivative is
f'(x) = (x^2+4)(2x -2) - (x^2-2x+4)(2x)]/(x^2 +4)^2
Set the numerator = 0 and solve for x.
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