009A Sample Final A, Problem 5

5. Consider the function   $h(x)={\displaystyle {\frac {x^{3}}{3}}-2x^{2}-5x+{\frac {35}{3}}}.$
   (a) Find the intervals where the function is increasing and decreasing.
   (b) Find the local maxima and minima.
   (c) Find the intervals on which $f(x)$ is concave upward and concave downward.
   (d) Find all inflection points.
   (e) Use the information in the above to sketch the graph of $f(x)$.

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 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:

Find the Derivatives and Roots:
Note that
$f'(x)=x^{2}-4x-5=(x-5)(x+1),$
while
$f''(x)=2x-4.$
The roots of f ' are -1 and 5, while the only root of f " is 2.

Produce Sign Charts and Evaluate:
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 -1, between -1 and 5 and above 5. For example:
$f'(-10)=(-)(-)=(+),\quad f'(0)=(-)(+)=(-),\quad f'(10)=(+)(+)=(+).$
From this, we can build a sign chart:
$x:$ $x<-1$ $x=-1$ $-1<x<5$ $x=5$ $x>5$ $f'(x):$ $(+)$ $0$ $(-)$ $0$ $(+)$
This gives us the following answers:
(a) The function is increasing on $-(\infty ,-1)$ and $(5,\infty )$, and decreasing on $(-1,5)$.
(b) The first derivative test shows
$f(-1)=-{\frac {1}{3}}-2+5+{\frac {35}{3}}=14\,{\frac {1}{3}}$
is a local maximum, while
$f(5)={\frac {125}{3}}-50-25+{\frac {35}{\,3}}=-75+{\frac {160}{3}}=-\,{\frac {65}{\,3}}.$
is a local minimum.
The second derivative has only the single root x = 2, so we need only look at values for x < 2 and x > 2. These values are clearly negative and positive, respectively, so we have a sign chart:
$x:$ $x<2$ $x=2$ $x>2$ $f''(x):$ $(-)$ $0$ $(+)$
This gives us the following answers:
(c) The function is concave downward on $(-\infty ,2)$ and concave upward on $(2,\infty )$.
(d) The function has an inflection point at $(2,f(2))=\left(2,-{\frac {11}{3}}\right).$

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