(x = + 2 C[1]+ Log 1 (1+ 2

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13 ||x = 2 C[1]+ Log 1 1 + 13 2

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6.8 Sketch the graphs of f(x) = x3 7 x2 + 2 x + 20 and g(x) = x2 on the same set of axes and find their points of intersection exactly and approximately.

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f[x_] = x3 7 x2 + 2 x + 20; g[x_] = x2; Plot[{f[x], g[x]},{x, 10, 10},PlotRange { 100, 100}]

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xvalues = Solve[f[x] g[x], x];

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f[x]} /. xvalues //Expand 19,28 - 6 19 } , {3 + 19,28 + 6 19 }

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{{2,4},{3

% // N

{{2.,

4.}, { 1.3589, 1.84661}, {7.3589, 54.1534}}

6.9 A theorem from algebra says that if p( x ) = an x n + an 1 x n 1 + an 2 x n 2 + L + a1 x + a0 , the sum of the a a roots of the equation p(x) = 0 is n 1 and their product is ( 1)n 0 . Verify this for the equation an an 20 x 7 + 32 x 6 221x 5 118 x 4 + 725 x 3 18 x 2 726 x + 252 = 0

SOLUTION

solution = Solve [20 x7 + 32 x6 221 x5 118 x4 + 725 x3 18 x2 726 x + 252 0] 2 {{x 7} ,{x 5} ,{x 3} ,{x 2} ,{x 2 2 list = x /. solution 2 { 7 , 5 , 3 , 2, 2 2 2, 3, 3 2 } , {x 3 } , {x 3 }

n = 7;

list[[k]]

8 5

or Sum[list[[k]], {k, 1, 7}] or Total[list] an 1 32 8 = = 20 5 an

list[[k]]

Product[list[[k]], {k, 1, 7}]

n = 7; ( 1)n

63 5

a0 252 63 = ( 1) 7 = 20 5 an

6.10 Find all possible solutions, x, for the equation ax + b = cx + d.

SOLUTION

Solve[a x + b c x + d, x]

{{x b +cd }} a

This solution presumes a c. A more general solution is obtained using Reduce. Reduce[a x + b c x + d, x] (b d && a c) || a c 0 && x b + d a c

6.11 Eliminate the variable x from the nonlinear system x 3 + y2 + z = 1 x+ y+z =3

Equations

SOLUTION

Eliminate[{x3 + y2 + z 1, x + y + z 3}, x] (26 18 y + 3 y2) z + ( 9 + 3 y) z2 + z3 26 27 y + 10 y2 y3

6.2 Solving Transcendental Equations

A transcendental equation is one that is non-algebraic. Although Solve and NSolve can be used in a limited way to handle simple trigonometric or exponential equations, it was not designed to handle equations involving more complicated transcendental functions. The Mathematica command FindRoot is better equipped to handle these. FindRoot uses iterative methods to find solutions. A starting value, sometimes called the initial guess, must be specified. For best results, the initial guess should be as close to the desired root as possible.

FindRoot[lhs rhs, {x, x0}] solves the equation lhs = rhs using Newton s method with starting value x0. FindRoot[lhs rhs,{x, {x0, x1}] solves the equation lhs = rhs using (a variation of) the secant method1 with starting values x0 and x1. FindRoot[lhs rhs,{x, x0, xmin, xmax}] attempts to solve the equation, but stops if the iteration goes outside the interval [xmin, xmax].

If a function is specified in place of the equation lhs rhs, FindRoot will compute a zero of the function. A zero of f is a number x such that f(x) = 0.

EXAMPLE 16 The equation sin x = x2 1 has two solutions.

Plot[{Sin[x], x2 1}, {x, o, o}]

4 3 2 1

1 1 2

The graph of the two functions shows that they intersect near x = 1 and x = 1. FindRoot[Sin[x] x2 1, {x, 1}]

0.636733} 1.40962}

FindRoot[Sin[x] x2 1, {x, 1}]

1 Newton s method uses the x-intercept of the tangent line to improve the accuracy of the initial guess. Thus, Newton s method fails if the derivative of the function cannot be computed. The secant method, although a bit slower, uses the values of the function at two distinct points, computing the x-intercept of the secant line.