Test 3 is April 26th and will cover the two projects on derivatives.
We will use functions provided by the MTH229 package:
using MTH229
We discussed three ways to find derivatives in Julia:
We can wrap this in a function, but perhaps it is best to just compute it. For example, for the function $f(x) = x - \log(x)$, we have the derivative at $2$ is approximately:
f(x) = x - log(x) c = 2 h = 1e-6 (f(c+h) - f(c)) / h # basically 1/2, but an approximation
Different values of h give different approximations and there is an art to selecting the right size of h, but for most purposes, 1e-6 will be reasonable.
h goes to zero by just using some small value of h. For most problems, the "automatic derivative" can be found and will be more accurate. The MTH229 package overrides the syntax for the ' to make computing these straightforward:f(x) = x - log(x) c = 2 f'(c)
Second derivatives are found by using 2 "primes":
f''(c)
We can use f' just like any other function gets used. For example, to find out when $f'(x) = 0$ in the interval $[1/10, 10]$ we have:
fzeros(f', 1/10, 10)
1-element Array{Float64,1}:
1.0
diff function,as in diff(f).At $c=2$, we can see the symbolic derivative is also $1/2$ with diff(f)(2).
The project introduced the tangent function to describe the tangent line to $f$ and $c$. We can plot this as follows:
f(x) = x^5 - x - 1 plot(f, 0, 2) c = 2 plot!(tangent(f, c), 0, 2)
We can ask questions about the tangent line. For example:
When will the tangent line cross the $x$ axis? Clearly it is between 0 and 2:
fzeros(tangent(f, c), 0, 2)
1-element Array{Real,1}:
1.63291
Or where will the tangent line cross the $y$ axis? This is the value at $0$:
tl = tangent(f, c) tl(0)
This rough table describes some relationships we explored in one of the projects:
| f | $+$ | $\nearrow$ | $\smile$ |
| f' | . | $+$ | $\nearrow$ |
| f'' | . | . | $+$ |
It helps us understand two "tests":
The first derivative test states that for a differentiable function $f(x)$ with a critical point at $c$ then if $f′(x)$ changes sign from + to − at c then $f(c)$ is a local maximum and if it changes sign from − to + then $f(c)$ is a local minimum
The second derivative test states that if $c$ is a critical point of $f(x)$ and $f″(c)>0$ then $f(c)$ is a local minimum and if $f″(c)<0$ then $f(c)$ is a local maximum.
Where a critical point is a value in the domain of $f(x)$ where the derivative is $0$ or undefined. In contrast, an inflection point is a point where $f''(x)$ is $0$ or undefined and the second derivative changes sign at this point.
To find these we used fzero and fzeros. The function fzero(f, a, b) will find a zero of $f(x)$ within a bracketing interval $[a,b]$. The function fzeros(f, a, b) will try to find the zeros of $f$ between $[a,b]$. (It may miss zeros that do not cross the $x$ axis!)
For example, if $f(x) = \cos(x) - \log(x)$ on $(0,10)$ we have the critical points found with:
f(x) = cos(x) - log(x) fzeros(f', 0, 10)
3-element Array{Float64,1}:
3.43683
6.11902
9.5299
(The function is differentiable on this interval.)
The inflection points are given by
fzeros(f'', 0, 10)
2-element Array{Float64,1}:
4.7566
7.8377
(which requires verifying the second derivative changes sign at these two points.)
This plotif function was helpful in graphical explorations:
For example, using the second derivative for f shows that indeed these two points above are inflection points
plotif(f, f'', 0, 10)
The hardest problems here deal with the case where a function is described in terms of its derivative (or second derivative). For example, suppose $f'(x) = x^3 - x + 1$. What can be said about where $f$ is increasing? Concave up?
Well, $f$ is increasing when $f'(x) \geq 0$ (basically), so we identify where $f$ is 0.
fp(x) = x^3 - x + 1 cps = fzeros(fp, -5, 5)
1-element Array{Real,1}:
-1.32472
We can graph or check otherwise that to the left of the lone zero $fp < 0$ and to the right $fp > 0$. So $f(x)$ will increase on $( -1.32472, \infty)$.
Now, $f$ will be concave up when $f''(x) \geq 0$. We can investigate this with:
fzeros(fp', -5, 5)
2-element Array{Float64,1}:
-0.57735
0.57735
We see two possible inflection points. A graph will show that $f''(x) < 0$ in $(-0.57735, 0.57735)$ and is otherwise positive. So $f(x)$ is concave up except on the interval $(-0.57735, 0.57735)$.