Problem: Sketch the graph of the function

f (x) = 3x⁴ − 8x³ + 6x².

Solution: We will use the procedure as outlined in Overview of graphing in Methods Survey - Graphing.

Step 1. The domain of this function is the whole real line, the function is continuous there. Since it is a polynomial with both even and odd powers, we suspect that it is not symmetric. We try one symmetric couple: For instance, f (1) = 1 while f (−1) = 17, which rules out any possibility that f is even or odd.

The x-intercept:

3x⁴ − 8x³ + 6x² = 0   iff   x²(3x² − 8x + 6) = 0   iff   x = 0.

There is only one intercept f (0) = 0, which is also the y-intercept.

Step 2. We find limits at endpoints of the interval given by the domain.

Asymptotes: Since the function is continuous on the whole real line, there cannot be vertical asymptotes. Because limits at negative infinity and infinity diverge, there are no horizontal asymptotes there, but the limits exist and therefore there is a chance for oblique asymptotes. We will use the appropriate algorithm.

The limit for A diverged at both negative and positive infinity, therefore there are no oblique asymptotes there.

Step 3. We find the derivative and use it to determine monotonicity and local extrema.

f ′(x) = 12x³ − 24x² + 12x = 12x(x² − 2x + 1) = 12x(x − 1)².

Critical points: There are no points in the domain where the derivative does not exist; the derivative is zero at x = 0 and at x = 1. The domain thus splits into three intervals of monotonicity, we determine it using a table; we put closed endpoints when the function is continuous.

We have adjacent intervals of equal monotonicity, and since f is continuous at the connecting point 1, we know that they can be connected. The conclusion is that f is decreasing on (−∞,0] and increasing on [0,∞).

Local extrema: The given function has a local minimum f (0) = 0. The point f (1) = 1 is not a local extreme.

Step 4. We find the second derivative and use it to determine concavity.

f ′′(x) = 36x² − 48x + 12 = 12(3x² − 4x + 1) = 12(3x − 1)(x − 1).

Dividing points: There are no points in the domain where the second derivative does not exist; the second derivative is zero at x = 1/3 and at x = 1. The domain thus splits into three intervals of concavity, we determine it using a table; we put closed endpoints when the function is continuous.

The conclusion is that f is concave up on (−∞,1/3] and on [1,∞); it is concave down on [1/3,1].

Inflection points: f (1/3) = 11/27 and f (1) = 1.

Step 5. Now we put it all together. First we put all points and limit trends that we obtained above into a picture. This will be the skeleton on which we will hang the function.

To see the shape of the graph better we combine the two tables above.

Now we are ready to sketch the graph. To make the picture more faithful we recall that at x = 1 there is a horizontal tangent line (the derivative was zero).

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