Graph 2x^3-3x^2-36x+5

Problem

2*x3−3*x2−36*x+5

Solution

1. Find the derivative to determine the critical points where the slope of the function is zero.

(d(2)*x3−3*x2−36*x+5)/d(x)=6*x2−6*x−36

2. Set the derivative to zero and solve for x to find the horizontal tangents.

6*(x2−x−6)=0

6*(x−3)*(x+2)=0

x=3,x=−2

3. Evaluate the original function at the critical points to find the local extrema.

ƒ(3)=2*(3)3−3*(3)2−36*(3)+5=−76

ƒ*(−2)=2*(−2)3−3*(−2)2−36*(−2)+5=49

4. Find the second derivative to determine the concavity and the inflection point.

(d(6)*x2−6*x−36)/d(x)=12*x−6

5. Set the second derivative to zero to find the point of inflection.

12*x−6=0

x=0.5

ƒ(0.5)=2*(0.5)3−3*(0.5)2−36*(0.5)+5=−13.5

6. Determine the end behavior by looking at the leading term 2*x3

As *x→∞,ƒ(x)→∞

As *x→−∞,ƒ(x)→−∞

7. Identify the y-intercept by setting x=0

ƒ(0)=5

Final Answer

To graph the function, plot the local maximum at (−2,49) the local minimum at (3,−76) the inflection point at (0.5,−13.5) and the y-intercept at (0,5) then connect them following the cubic end behavior.

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