Find the Local Maxima and Minima f(x)=(x+3)/(x-3)

Problem

ƒ(x)=(x+3)/(x−3)

Solution

1. Find the derivative using the quotient rule d()/d(x)u/v=(vd(u)/d(x)−ud(v)/d(x))/(v2)

ƒ(x)′=((x−3)d(x+3)/d(x)−(x+3)d(x−3)/d(x))/((x−3)2)

2. Simplify the numerator by evaluating the derivatives of the linear terms.

ƒ(x)′=((x−3)*(1)−(x+3)*(1))/((x−3)2)

ƒ(x)′=(x−3−x−3)/((x−3)2)

ƒ(x)′=(−6)/((x−3)2)

3. Identify critical points by setting the derivative equal to zero or finding where it is undefined.

(−6)/((x−3)2)=0

4. Analyze the results to determine if any local extrema exist. The equation (−6)/((x−3)2)=0 has no solution because a fraction is only zero if its numerator is zero, and −6≠0 The derivative is undefined at x=3 but x=3 is not in the domain of the original function ƒ(x) because it causes division by zero.

5. Conclude that since there are no critical points within the domain of the function, there are no local maxima or minima.

Final Answer

Local Maxima: None, Local Minima: None

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