Solution

Solve can not in general solve non-polynomial equations. It might give a false answer, or at least leave some solutions out. It will give a warning message when that might be the case.

In[69]:=

(*1*)Solve[Exp[x] 2, x] Solve[Sqrt[x * x - 1] 1/2, x] ᡝ ... 62513;2}, {x, y}] Solve[x Exp[x] 1, x] Solve[x Exp[Exp[x]] 1, x]

Solve :: ifun : Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. More…

Out[69]=

{{xLog[2]}}

Out[70]=

{{x -5^(1/2)/2}, {x5^(1/2)/2}}

Solve :: ifun : Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. More…

Out[71]=

{{x1, y1}, {x2, y0}}

InverseFunction :: ifun : Inverse functions are being used. Values may be lost for multivalued inverses. More…

Solve :: ifun : Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. More…

Out[72]=

{{xProductLog[1]}}

Solve :: tdep : The equations appear to involve the variables to be solved for in an essentially non-algebraic way. More…

Out[73]=

Solve[^^x x1, x]

Note that the last equation couldn't be solved at all. See the help file for ProductLog if you wish.

In[74]:=

(*2*)Reduce[Sin[x]^2 + Cos[x] 0, x] Reduce[Exp[2x] + 2Exp[x] - 30, x]

Out[74]=

C[1] ∈Integers&& (x -2  ArcTanh[(-2 + 5^(1/2))^(1/2)] + 2 ` ... [(2 + 5^(1/2))^(1/2)] + 2 π C[1] || x2 ArcTan[(2 + 5^(1/2))^(1/2)] + 2 π C[1])

Out[75]=

C[1] ∈Integers&& (x2  π C[1] || x π + 2  π C[1] + Log[3])

Reduce never makes approximations. The output corresponds with logical exactness to the input.

In[76]:=

(*3a*)Plot[Cos[x] - x, {x, -8, 8}] FindRoot[Cos[x] x, {x, 0.5}]

[Graphics:../HTMLFiles/ex02_solutions_188.gif]

Out[76]=

⁃Graphics⁃

Out[77]=

{x0.739085}

In[78]:=

(*3b*)Plot[Log[x] - 1/x^2, {x, 0.5, 3}] FindRoot[Log[x] 1/x^2, {x, 1.5}]

[Graphics:../HTMLFiles/ex02_solutions_192.gif]

Out[78]=

⁃Graphics⁃

Out[79]=

{x1.53158}

In 3.c), since the interval is allready given, we wouldn't have to plot the function to find the initial guess, but let's plot it anyway.

In[80]:=

(*3c*)Plot[BesselJ[0, x], {x, -10, 10}] FindRoot[BesselJ[0, x] 0, {x, 5, 6}]

[Graphics:../HTMLFiles/ex02_solutions_196.gif]

Out[80]=

⁃Graphics⁃

Out[81]=

{x5.52008}

The following function has infinitely many zeros, which you can see by increasing the range of x. Let's find the smallest root.

In[82]:=

(*3d*)Clear[x] g[x_] = Exp[x Sin[x]] - x ; Plot[g[x], {x, -8, 8}, PlotRange {-8, 12}] FindRoot[g[x] 0, {x, 2.5}]

[Graphics:../HTMLFiles/ex02_solutions_200.gif]

Out[84]=

⁃Graphics⁃

Out[85]=

{x2.76492}

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