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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