Graph f(x)=x^3-12x+1

Problem

ƒ(x)=x3−12*x+1

Solution

1. Find the y-intercept by evaluating the function at x=0

ƒ(0)=(0)3−12*(0)+1=1

The y-intercept is at (0,1)

2. Find the first derivative to determine the critical points and intervals of increase or decrease.

d(ƒ(x))/d(x)=3*x2−12

3. Solve for critical points by setting the first derivative equal to zero.

3*x2−12=0

3*(x2−4)=0

x=2,x=−2

4. Determine local extrema by evaluating the function at the critical points.

ƒ(2)=(2)3−12*(2)+1=8−24+1=−15

ƒ*(−2)=(−2)3−12*(−2)+1=−8+24+1=17

The local maximum is at (−2,17) and the local minimum is at (2,−15)

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

d2(ƒ(x))/(d(x)2)=6*x

6. Identify the inflection point by setting the second derivative to zero.

6*x=0

x=0

The inflection point is at (0,1) where the graph changes from concave down (x<0 to concave up (x>0.

7. Analyze end behavior of the cubic function.
   As x→∞ ƒ(x)→∞
   As x→−∞ ƒ(x)→−∞

Final Answer

To graph ƒ(x)=x3−12*x+1 plot the local maximum at (−2,17) the y-intercept and inflection point at (0,1) and the local minimum at (2,−15) then connect them with a smooth cubic curve.

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