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f (x) is continuous and differentiable everywhere. Define a = 0, b = and solve the equation f (b) f (a) = f ' (c)(b a) for c. To approximate their values, we look at the graph with the endpoints connected by a line segment. f[x_] = x + Sin[2 x] ; a = 0; b = o; m = f[b] f[a]; b a l[x_] = f[a]+ m (x a);
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Slope of the secant connecting the endpoints. Function representing the secant line.
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Differential Calculus
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Plot[{f[x], l[x]}, {x, a, b}]
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It looks like the tangent line will be parallel to the secant when x 1 or x 2.5. Clearly both values lie between 0 and . FindRoot[f[b] f[a] f'[c](b a), {c, 1}] FindRoot[f[b] f[a] f'[c](b a), {c, 2.5}]
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8.3 Maximum and Minimum Values
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A function f has an absolute (global) maximum over an interval, I, at a point c if f(x) f(c) for all x in I. In other words, f(c) is the largest value of f(x) in I. A similar definition (with the inequality reversed) holds for an absolute minimum. One of the most important applications of differential calculus is optimization, i.e., finding the maximum and minimum values of a function, subject to certain constraints. Not all functions have absolute maxima and minima. However the Extreme Value Theorem gives conditions sufficient to guarantee their existence: If f is continuous on a closed bounded interval, then f has both an absolute maximum and an absolute minimum in that interval. A critical number of a function f is a number c for which f '(c) = 0 or f '(c) fails to exist. It can be shown that if a function is continuous on the closed interval [a, b], then the absolute maximum and minimum will be found either at a critical number or at an endpoint of the interval. We can use Mathematica to help us find the maximum and/or minimum values.
EXAMPLE 15 We wish to find the absolute maximum and minimum values of the function f (x) = x4 4 x3 + 2 x2 + 4 x + 2 on the interval [0, 4]. First we find the critical numbers.
f[x_]= x4 4 x3 + 2 x2 + 4 x + 2; Solve[f'[x] 0]
{{x 1},{x 1 2},{x 1 + 2}}
Of these three numbers, only two lie in the interval [0, 4]. We compute the value of the function at these numbers as well as the endpoints of the interval. c1 = 0; c2 = 1; c3 = 1 + 2; c4 = 4; points = {{c1, f[c1]}, {c2, f[c2]}, {c3, f[c3]}, {c4, f[c4]}} //Expand;
Differential Calculus
TableForm[points, TableHeadings {None, {"x", "f[x]"}}]
x 0 1 1+ 2 4 f[x] 2 5 1 50
Max[{f[c1], f[c2], f[c3], f[c4]}] //Expand 50 Min[{f[c1], f[c2], f[c3], f[c4]}] //Expand 1 The absolute maximum of f is 50 and the absolute minimum is 1.
EXAMPLE 16 A wire, 100 in. long, is to be used to form a square and a circle. Determine how the wire should be distributed in order for the combined area of the two figures to be (a) as large as possible and (b) as small as possible.
100" x
The combined area of the two figures is A( x ) = x 2 + r 2. The circle has a circumference of 2 r, so it follows that 4 x + 2 r = 100 . Since the wire is 100 in. long, 0 x 25. Solve[4 x + 2 o r 100, r]
r 2( 25 + x)
Replace r in terms of x.
a[x_]= x2 + o r2 /. r 2( 25 + x) o
2 ) 4( 25 + x) + x2
Solve[a'[x] 0] 100 x 4+ x1 = 0; x2 = 100 ; 4+ o x3 = 25;
Find critical value(s).
Compute the values of a(x) at these three points. The values are placed in a table with numerical approximations for comparison.
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