Maxima, Minima, and Inflection Points

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This demonstration shows how to find extrema of functions using analytical and numerical techniques using the Symbolic Math Toolbox™.

First Derivatives: Finding Local Minimum and Maximum of the Function
Second Derivatives: Finding Inflation Points of the Function
Limits: Functions with Suprema

First Derivatives: Finding Local Minima and Maxima

Computing the first derivative of an expression helps you find local minima and maxima of that expression. For example, create a rational expression where the numerator and the denominator are polynomial expressions:

symsxf = (3 * x^3 + 17 * x^2 + 6 * x + 1)/(2 * x^3 + x * -1 + 3)

f =
                  
                    $\frac{3 x^{3} + 17 x^{2} + 6 x + 1}{2 x^{3} - x + 3}$

Plotting this expression shows that it has horizontal and vertical asymptotes, a local minimum between -1 and 0, and a local maximum between 1 and 2:

fplot(f)

Figure contains an axes object. The axes object contains an object of type functionline.

By default, when you operate on this expression, results can include both real and imaginary numbers. If you are interested in real numbers only, you can set the permanent assumption that x belongs to the set of real numbers. This allows you to avoid complex numbers in the solutions and it also can improve performance:

assume(x,'real')

To find a horizontal asymptote, compute the limit of $f$ for x approaching positive and negative infinities. The horizontal asymptote is $x = 3 / 2$ :

[limit(f, x, sym(inf)), limit(f, x, -sym(inf))]

ans =
                  
                    $(\begin{array}{c} \frac{3}{2} & \frac{3}{2} \end{array})$

To find a vertical asymptote of $f$ , find the roots of the polynomial expression that represents the denominator of $f$ :

solve(2 * x^3 + x * -1 + 3 == sym(0), x)

ans =
                  
                    $root (z^{3} - \frac{z}{2} + \frac{3}{2}, z, 3)$

To get an explicit solution for such equations, try calling the solver with the optionMaxDegree. The option specifies the maximum degree of polynomials for which the solver tries to return explicit solutions. By default,MaxDegree= 2. Increasing this value, you can get explicit solutions for higher-order polynomials. For example, specifyingMaxDegree= 3results in an explicit solution:

solve(2 * x^3 + x * -1 + 3 == 0, x,'MaxDegree', 3)

ans =
                  
                    $- \frac{1}{6 {(\frac{3}{4} - \frac{\sqrt{241} \sqrt{432}}{432})}^{1 / 3}} - {(\frac{3}{4} - \frac{\sqrt{241} \sqrt{432}}{432})}^{1 / 3}$

You can approximate the exact solution numerically by using thevpafunction.

vpa(ans,6)

ans =
                   $- 1.28962$

Now find the local minimum and maximum of the expressionf. If the point is a local extremum (either minimum or maximum), the first derivative of the expression at that point is equal to zero. To compute the derivative of an expression, use thedifffunction:

g = diff(f, x)

g =
                  
                    $\frac{9 x^{2} + 34 x + 6}{2 x^{3} - x + 3} - \frac{(6 x^{2} - 1) (3 x^{3} + 17 x^{2} + 6 x + 1)}{{(2 x^{3} - x + 3)}^{2}}$

To find the local extrema off, solve the equationg = 0. If you use theMaxDegreeoption, the solver returns the long explicit solution, which can be approximated by using thefloatfunction:

solve(g == 0, x,'MaxDegree', 4); extrema = vpa(ans, 6)

extrema =
                  
                    $(\begin{array}{c} - 0.189245 \\ 1.28598 \end{array})$

The plot of the expressionfshows that x = -0.189 is a local minimum of the expression, and x = 1.286 is its local maximum.

fplot(f) holdonplot(extrema, subs(f,extrema),'*') holdoff

Figure contains an axes object. The axes object contains 2 objects of type functionline, line.

Second Derivatives: Finding Inflection Points

Computing the second derivative lets you find inflection points of the expression.

h(x) = simplify(diff(f, x, 2))

h(x) =
                  
                    $\frac{2 (68 x^{6} + 90 x^{5} + 18 x^{4} - 699 x^{3} - 249 x^{2} + 63 x + 172)}{{(2 x^{3} - x + 3)}^{3}}$

To find inflection points of $f$ , solve the equationh = 0. For this equation the symbolic solver returns a complicated result even if you use theMaxDegreeoption:

solve(h == 0, x,'MaxDegree', 4)

ans =
                  
                    $\begin{array}{l} (\begin{array}{c} root (σ_{1}, z, 1) \\ root (σ_{1}, z, 4) \end{array}) \\ where \\ σ_{1} = z^{6} + \frac{45 z^{5}}{34} + \frac{9 z^{4}}{34} - \frac{699 z^{3}}{68} - \frac{249 z^{2}}{68} + \frac{63 z}{68} + \frac{43}{17} \end{array}$

To get the simpler numerical result, solve the equation numerically by usingvpasolve; specify the search range to restrict the returned results to all real solutions of the expression:

inflection = vpasolve(h == 0, x, [-inf, inf])

inflection =
                  
                    $(\begin{array}{c} 0.57871842655441748319601085860196 \\ 1.8651543689917122385037075917613 \end{array})$

The expressionfhas two inflation points: x = 0.579 and x = 1.865.

fplot(f) holdonplot(extrema, subs(f,extrema),'*') plot(inflection, subs(f,inflection),'*') holdoff

Figure contains an axes object. The axes object contains 3 objects of type functionline, line.

Suprema

Not all functions can be treated analytically; the function

$f (x) = t a n (s i n (x)) - s i n (t a n (x))$

is very flat at the origin and it oscillates infinitely often near $- \frac{π}{2}$ , becomes linear as it approaches zero and oscillates again near $π$ .

f = @(x) tan(sin(x))-sin(tan(x))

f =function_handle with value:@(x)tan(sin(x))-sin(tan(x))

fplot(f, [-pi , pi])

Most important for our purposes here, fplot has picked the limit on they-axes to be

ylim

ans =1×2-2.5488 2.5572

What is happening at $\frac{π}{2}$ ?

MATLAB® uses double precision arithmetic, so $\frac{π}{2}$ evaluates to one of the oscillations.

pi/2

ans = 1.5708

f(pi/2)

ans = 2.5161

The Symbolic Math Toolbox uses exact arithmetic, which shows the function is undefined.

symsxsym(pi/2)

ans =
                  
                    $\frac{π}{2}$

F = sym(f)

F =
                   $\tan (\sin (x)) - \sin (\tan (x))$

subs(F,x,sym(pi/2))

ans =
                   $NaN$

Taylor Series

We can also try to look at the value with a Taylor Series.

T = taylor(F,x,'Order',10,'ExpansionPoint',0)

T =
                  
                    $\frac{29 x^{9}}{756} + \frac{x^{7}}{30}$

vpa(subs(T,x,pi/2))

ans =
                   $3.0198759869735883213825972535797$

holdonfplot(T) ylim ([-3 3]) holdoff

Figure contains an axes object. The axes object contains 2 objects of type functionline.

Calculus

We learn in calculus that a maximum occurs at a zero of the derivative. But this function is not differentiable in the vicinity of $\frac{π}{2}$ . We can analytically differentiate $f (x)$ using the Symbolic Math Toolbox.

diff(F)

ans =
                   $\cos (x) ({\tan (\sin (x))}^{2} + 1) - \cos (\tan (x)) ({\tan (x)}^{2} + 1)$

fplot(diff(F), [-pi , pi])

Figure contains an axes object. The axes object contains an object of type functionline.

Sampling

We can sample the function N times near $\frac{π}{2}$ to get a numerical approximation to the value of the maximum. Is that good enough?

N = 100; xValues = 3*pi/8 + pi/4*rand(1,N)

xValues =1×1001.8180 1.8895 1.2778 1.8955 1.6748 1.2547 1.3968 1.6076 1.9301 1.9359 1.3019 1.9404 1.9299 1.5593 1.8066 1.2895 1.5093 1.8973 1.8003 1.9317 1.6931 1.2061 1.8450 1.9117 1.7112 1.7732 1.7618 1.4862 1.6929 1.3125 1.7326 1.2031 1.3956 1.2144 1.2544 1.8248 1.7238 1.4271 1.9244 1.2052 1.5227 1.4778 1.7793 1.8026 1.3249 1.5628 1.5281 1.6857 1.7352 1.7708

ySoln = f(xValues)

ySoln =1×1000.7260 1.5080 1.5932 1.5614 1.3796 1.3158 2.0658 - 2.4586 1.8194 1.8541 1.9088 1.8793 1.8178 2.3439 0.6145 1.7447 2.0697 1.5775 0.5629 1.8290 2.4930 0.8543 1.0347 1.6931 2.2371 0.5024 0.6011 2.2489 2.4891 2.0499 1.3564 0.8308 2.0986 0.9208 1.3122 0.8011 1.7177 0.9333 1.7830 0.8466 0.6138 2.5047 0.4857 0.5809 2.2051 2.5133 2.5387 2.2247 1.2583 0.5153

max(ySoln)

ans = 2.5387

Proof

Determine the maximum from a mathematical proof.

$\sin (x) \leq 1$

so $\sin (\tan (x)) \leq 1$ and $t a n (s i n (x)) \leq \tan (1)$ which means consequently

$f (x) \leq 1 + t a n (1)$

As $x \Rightarrow \frac{π}{2}$ , $\tan (x)$ oscillates and blows up; so $f (x)$ is actually not defined at all at this point as was shown above

$f (x) < 1 + \tan (1)$

Now we can take a look at the numerical value.

formatlong1 + tan(1)

ans = 2.557407724654902