Find the Local Maxima and Minima y=x^3-4x^2-16x+3

Problem

y=x3−4*x2−16*x+3

Solution

1. Find the first derivative of the function to determine the slope of the tangent line.

d(y)/d(x)=3*x2−8*x−16

2. Set the derivative to zero to find the critical points where the slope is horizontal.

3*x2−8*x−16=0

3. Solve the quadratic equation using the quadratic formula x=(−b±√(,b2−4*a*c))/(2*a)

x=(8±√(,(−8)2−4*(3)*(−16)))/(2*(3))

x=(8±√(,64+192))/6

x=(8±√(,256))/6

x=(8±16)/6

(x_1)=4

(x_2)=−4/3

4. Find the second derivative to apply the Second Derivative Test for concavity.

d2(y)/(d(x)2)=6*x−8

5. Test the critical points by substituting them into the second derivative.

For *x=4:6*(4)−8=16>0⇒Local Minimum

For *x=−4/3:6*(−4/3)−8=−16<0⇒Local Maximum

6. Calculate the y-coordinates by substituting the critical points back into the original function.

y(4)=(4)3−4*(4)2−16*(4)+3=−61

y*(−4/3)=(−4/3)3−4*(−4/3)2−16*(−4/3)+3=445/27

Final Answer

Local Maximum: *(−4/3,445/27), Local Minimum: *(4,−61)

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