Graphics Programs Reference
In-Depth Information
intersection we will have to use three different starting points. The graphical
analysis above suggests appropriate starting points.
r1 = fzero(inline('2ˆx - xˆ4'), -0.9)
r2 = fzero(inline('2ˆx - xˆ4'), 1.2)
r3 = fzero(inline('2ˆx - xˆ4'), 16)
r1 =
-0.8613
r2 =
1.2396
r3 =
16
Let's check that these “solutions” satisfy the equation.
subs('2ˆx - xˆ4', 'x', r1)
subs('2ˆx - xˆ4', 'x', r2)
subs('2ˆx - xˆ4', 'x', r3)
ans =
2.2204e-016
ans =
-8.8818e-016
ans =
0
So
r1
and
r2
very nearly satisfy the equation, and
r3
satisfies it exactly. It is
easily seen that 16 is a solution. It is also interesting to try
solve
on this
equation.
symroots = solve('2ˆx - xˆ4 = 0')
symroots =
[ -4*lambertw(-1/4*log(2))/log(2)]
[ 16]
[ -4*lambertw(-1/4*i*log(2))/log(2)]
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