Find the Local Maxima and Minima y=(x^2-5)/(x-3)

Problem

y=(x2−5)/(x−3)

Solution

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

d(y)/d(x)=((x−3)d(x2−5)/d(x)−(x2−5)d(x−3)/d(x))/((x−3)2)

2. Differentiate the numerator terms.

d(y)/d(x)=((x−3)*(2*x)−(x2−5)*(1))/((x−3)2)

3. Simplify the numerator to find the expression for the derivative.

d(y)/d(x)=(2*x2−6*x−x2+5)/((x−3)2)

d(y)/d(x)=(x2−6*x+5)/((x−3)2)

4. Identify critical points by setting the derivative equal to zero.

x2−6*x+5=0

(x−5)*(x−1)=0

x=5,x=1

5. Apply the second derivative test or analyze the sign of the first derivative. We find the second derivative.

d2(y)/(d(x)2)=((x−3)2*(2*x−6)−(x2−6*x+5)*(2*(x−3)))/((x−3)4)

d2(y)/(d(x)2)=(2*(x−3)2−2*(x2−6*x+5))/((x−3)3)

d2(y)/(d(x)2)=(2*(x2−6*x+9)−2*x2+12*x−10)/((x−3)3)

d2(y)/(d(x)2)=8/((x−3)3)

6. Evaluate the second derivative at the critical points to determine the nature of the extrema.

At *x=1:d2(y)/(d(x)2)=8/((1−3)3)=−1<0⇒Local Maxima

At *x=5:d2(y)/(d(x)2)=8/((5−3)3)=1>0⇒Local Minima

7. Calculate the y-coordinates for the local maxima and minima.

y(1)=(1−5)/(1−3)=(−4)/(−2)=2

y(5)=(5−5)/(5−3)=20/2=10

Final Answer

Local Maxima: *(1,2), Local Minima: *(5,10)

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