# 022 Sample Final A, Problem 6

Sketch the curve, including all relative extrema and points of inflection: ${\displaystyle y=3x^{4}-4x^{3}}$.

Foundations:
We learn a lot about the shape of a function's graph from its derivatives. When a first derivative is positive, the function is increasing (heading uphill). When the first derivative is negative, it is decreasing (heading downhill). Of particular interest is when the first derivative at a point is zero. If f '(z) = 0 at a point z, and the first derivative splits around it (either f '(x) < 0 for x < z and f '(x) > 0 for x > z or f '(x) > 0 for x < z and f '(x) < 0 for x > z), then the point (z,f(z)) is a local maximum or minimum, respectively.
The second derivative tells us how the first derivative is changing. If the second derivative is positive, the first derivative (the slope of the tangent line) is increasing. This is equivalent to the graph "turning left" if we consider moving from negative x-values to positive. We call this "concave up". The parabola y = x2 is an example of a purely concave up graph, and its second derivative is the constant function y " = 2.
If the second derivative is negative, then the first derivative is decreasing. This means we are turning right as we move from negative x-values to positive. This is called "concave down". The inverted parabola y = -x2 is an example of a purely concave down graph.
A point z where the second derivative is zero, and the sign of the second derivative splits around it (either f "(x) < 0 for x < z and f "(x) > 0 for x > z, or f "(x) > 0 for x < z and f "(x) < 0 for x > z), then the point (z,f(z) is an inflection point.

Of course, there are tests we use to find local extrema (maxima and minima, which is the plural of maximum and minimum). We are assuming the function f is continuous and differentiable in an interval containing the point x0.
First Derivative Test: If at a point x0, f '(x0) = 0, and f '(x) < 0 for x < x0 while f '(x) > 0 for x > x0, then f(x0) is a local minimum.
On the other hand, if f '(x0) = 0, and f '(x) > 0 for x < x0 while f '(x) < 0 for x > x0, then f(x0) is a local maximum.
Second Derivative Test: If at a point x0, f '(x0) = 0, and f "(x0) > 0, then f(x0) is a local minimum. On the other hand, if f "(x0) < 0, then f(x0) is a local maximum. If f "(x0) = 0, the test is inconclusive.

Solution:

Step 1:
Find the Derivatives and Their Roots. Note that
${\displaystyle y'\,=\,12x^{3}-12x^{2}\,=\,12x^{2}(x-1).}$
This has roots ${\displaystyle 0}$ and ${\displaystyle 1}$.
On the other hand,
${\displaystyle y''\,=,36x^{2}-24x\,=\,12x(3x-2).}$
This has roots ${\displaystyle 0}$ and ${\displaystyle 2/3}$.
For graphing, I would also note that ${\displaystyle y=x^{3}(3x-4)}$, which has roots ${\displaystyle 0}$ and ${\displaystyle 4/3}$.
Step 2:
Produce Sign Charts and Evaluate. It should be clear that the function we're graphing is "roughly" like ${\displaystyle x^{4}}$, which should tell you it goes to positive infinity off to the left and right.

Since all of our tests rely on the signs of our derivatives, we need to produce sign charts. For the first derivative, we can test values below ${\displaystyle 0}$, between ${\displaystyle 0}$ and ${\displaystyle 1}$ and above ${\displaystyle 1}$. Using the factored version of each derivative, we can evaluate quickly. For example:

${\displaystyle f'(-10)=(+)(-)=(-),\quad f'(1/2)=(+)(-)=(-),\quad f'(10)=(+)(+)=(+).}$
From this, we can build a sign chart:
 ${\displaystyle x:}$ ${\displaystyle x<0}$ ${\displaystyle x=0}$ ${\displaystyle 0 ${\displaystyle x=1}$ ${\displaystyle x>1}$ ${\displaystyle y':}$ ${\displaystyle (-)}$ ${\displaystyle 0}$ ${\displaystyle (-)}$ ${\displaystyle 0}$ ${\displaystyle (+)}$
This tells us the point at the origin is not a local extrema, as the first derivative does not split around it. On the other hand, by the first derivative test we have a local minimum at ${\displaystyle (1,-1)}$.
The process is similar for the second derivative. Here, we can test values below ${\displaystyle 0}$, between ${\displaystyle 0}$ and ${\displaystyle 2/3}$ and above ${\displaystyle 2/3}$. Using the factored version of each derivative, we evaluate test points:
${\displaystyle f'(-10)=(-)(-)=(+),\quad f'(1/2)=(+)(-)=(-),\quad f'(10)=(+)(+)=(+).}$
From this, we can build a sign chart:
 ${\displaystyle x:}$ ${\displaystyle x<0}$ ${\displaystyle x=0}$ ${\displaystyle 0 ${\displaystyle x=2/3}$ ${\displaystyle x>2/3}$ ${\displaystyle y'':}$ ${\displaystyle (+)}$ ${\displaystyle 0}$ ${\displaystyle (-)}$ ${\displaystyle 0}$ ${\displaystyle (+)}$
This tells us the function is concave downward when ${\displaystyle 0, and concave upwards everywhere else. Additionally, there is an inflection point at ${\displaystyle \left({\frac {2}{3}},{\frac {16}{81}}\right)}$, although a teacher may not require you to compute the actual ${\displaystyle y}$-value.
Step 3:
Graph: The graph is drawn in blue where it is concave downward, and in red where it is concave upward. You should probably label the local minimum and inflection point, if you have time.