# Thread: 2nd Derivative Test and sketching the curve

1. ## 2nd Derivative Test and sketching the curve

Could someone walk me through this problem to the answer. I somewhat understand the first step but get lost when I have to use the 2nd derivative test and sketch the curve.

Find any critical numbers and then use the 2nd derivative test to decide whether the critical numbers lead to a relative maxima or minima. Sketch the curve
f(x) = x^3 - 2x^2 - 4x + 5

2. Originally Posted by Mathamateur
Could someone walk me through this problem to the answer. I somewhat understand the first step but get lost when I have to use the 2nd derivative test and sketch the curve.

Find any critical numbers and then use the 2nd derivative test to decide whether the critical numbers lead to a relative maxima or minima. Sketch the curve
f(x) = x^3 - 2x^2 - 4x + 5
Let's see, if I have time.

Critical numbers here are values of x when f'(x) = 0. They are where the graph of the function is steady; or not increasing, not decreasing; or where the tangent line is horizontal.

f'(x) = 3x^2 -4x -4
Set that to zero,
0 = 3x^2 -4x -4.
(3x +2)(x -2) = 0
x = -2/3 or 2 --------------at critical points.

2nd derivative test is about change in slope. If the slope of the tangent line changes from positive to negative, the critical point is maximum. From negative to positive f'(x), the critical point is minimum.

Or, in other words, if the 2nd derivative at the critical point is negative, then the critical point is maximum. Positive 2nd derivative means the critical point is minimum.

Or, 2nd deivative is about concavity of the graph also. If the concavity of the curve at the critical point if facing or opens downward, then that is for maximum. Concavity opening upward is for minimum.

Blah, blah, blah.
Words!

Anyway, the 2nd derivative, f''(x) = d/dx f'(x):
f'(x) = 3x^2 -4x -4
f''(x) = 6x -4

Test the critical mpoint at x = -2/3:
f''(-2/3) = 6(-2/3) -4 = -4 -4 = -8 -----negative.
Therefore, the graph has a maximum at x = -2/3 -------------***

Test the critical mpoint at x = 2:
f''(-2/3) = 6(2) -4 = 12 -4 = 8 -----positive.
Therefore, the graph has a minimum at x = 2 -------------***

Gotta go.

3. Ok, thanks. I think I understand so far.

But wouldn't the maxima and minima be reversed? x = 2 being the maximum and x = -2/3 being the minumum?

Either way..how would I go about sketching the curve now?

If someone could jump in since ticbol had to leave that would be great.

4. Originally Posted by Mathamateur
Could someone walk me through this problem to the answer. I somewhat understand the first step but get lost when I have to use the 2nd derivative test and sketch the curve.

Find any critical numbers and then use the 2nd derivative test to decide whether the critical numbers lead to a relative maxima or minima. Sketch the curve
f(x) = x^3 - 2x^2 - 4x + 5
f'(x)=3x^2-4x-4,

so the critical points are the roots of f'(x)=0, which we find using the quadratic formula to be x=2, -2/3.

Now f''(x)=6x-4, so f''(2)=8>0, so x=2 is a local minimum, and f''(-2/3)=-8<0,
so x=-2/3 is a local maximum.

RonL

5. Ok, so how would I go about sketching the curve with that information?

6. Originally Posted by Mathamateur
Ok, thanks. I think I understand so far.

But wouldn't the maxima and minima be reversed? x = 2 being the maximum and x = -2/3 being the minumum?

Either way..how would I go about sketching the curve now?

If someone could jump in since ticbol had to leave that would be great.
Okay, let's continue.

x = 2 being at the maximum and x = -2/3 being at the minimum.

Why, because 2 is greater than -2/3?
No, that is not how it goes. Remember, x = -2/3 is not the maximum point. It is the value of x where there is a maximum point. It is at that x where the graph has a maximum point.
[A graph may have many maximum points, or maxima, or relative maxima, or local maxima. The highest among them all, in the given interval or domain of the variable (where the function is continuous), x here, is the absolute maximum in that interval--if the y at both ends of the interval are lower than that highest maximum (Why, if any y at the ends of the interval is higher than that highest maximum inside, then the higher y is the absolute maximum in that interval. Blah, blah.) The rest are relative or local maxima. Same with minima.]

To see if x=2 is where there is a minimum, and if x = -2/3 is where there is a maximum, get the y at x=2 and the y at x = -2/3. Or, get the f(2) and f(-2/3).

f(x) = x^3 -2x^2 -4x +5

f(2) = 2^3 -2(2^2) -4(2) +5 = -7 ----------------------- y = -7, at x=2.
f(-2/3) = (-2/3)^3 -2(-2/3)^2 -4(-2/3) +5 = 6.48... ----- y = about 6.48, at x = -2/3.
See?
f(-2/3) is greater than f(2).

-----------------------------------------
To graph y = f(x) = x^3 -2x^2 -4x +5.

f(x) is cubic polynomial in x. That means its graph is a vertical S-curve, where, from left to right, it starts at after negative infinity, goes upward like crazy, reaches a maximum, goes downward, reaches a minimum, goes upward again like crazy until it apporaches positive infinity. Vertical S.

Suppose you don't know that. That x^3 something is vertical S. That's okay. Still you can graph the cubic curve, knowing the two critical points.

Draw the x,y axes. And the vertical lines x = -2/3 and x = 2. Plot the critical points (-2/3,6.48) and (2,-7).

Then get at least 3 more points. One to the left of x = -2/3. One at x=0 or the y-axis. One to the right of x=2. If you like, get a 4th point, at y=0.

If you like, if you want a more true graph, then get 2 points (far apart, say at x = -10, and then at x = -2) to the left of x = -2/3. Get the inflection point--where the concavity changes from facing down to facing up, where f"(x) is zer0. And get two pints to the right of x=2 (say at x=4 and at x=8).

Plot those points. Then connect those points with a "smooth" curve, knowing that as the curve approaches the maximum point at (-2/3,6.48), the curve bends to go flat/horizontal at that maximum point. Then it goes down. It changes concavity somewhere--at the inflection point--then it goes down still, flatenning again at the minimum point (2,-7). Then slowly concaving upwards again until it shoots up to the sky like crazy.