HP 40gs HP 39gs_40gs_Mastering The Graphing Calculator_English_E_F2224-90010.p - Page 107

Multiple solutions and the initial guess, Example 1, Graphing in Solve

Page 107 highlights

Multiple solutions and the initial guess Our first example was fairly simple because there was only one solution so it did not much matter where we began looking for it. When there is more than one possible answer you are required to supply an initial estimate or guess. The Solve aplet will then try to find a solution which is 'near' to the estimate. Example 1 The volume of a cylinder is given by V = 2π r (r + h) . Find the radius of a cylinder which has a volume of 1 liter and a height of 10cm. Enter the equation into E1 as shown right. When you are entering the equation, ensure that you put a * sign between the R and the bracket. See page 79 for more information on the reason for this. Change to the NUM view and enter the known values, remembering that 1 liter=1000cm3. Position the highlight over R, enter a positive value as your estimate, and press to find the solution shown right of 8.57cm. The equation is a quadratic in R which means two solutions are possible. If you enter an initial estimate of -10 you will obtain the negative solution, which is physically invalid in this case. Example 2 If f ( x) = x3 − 2x2 − 5x + 2 find all values of x for which f ( x) = 1. Although you may have a clear picture in your mind and can provide Solve with the estimates it needs, I'll assume that, like me, you would find it helpful to see a graph first. It is also possible to solve this in the Function aplet, which offers more powerful tools. The PLOT view in the Solve aplet, although powerful, can be deceptive if you don't understand it and I sometimes find it easier to work in the Function aplet. In this case we will continue to work in Solve. Graphing in Solve In the SYMB view, enter the equation Y=X^3-2X2-5X+2 into E1. In the NUM view, enter the known value of Y=1, ensure that the highlight is on X, making it the active variable, and then press PLOT. If your view does not look like this then you may not have had the highlight on the X, or your axes may not be set the same as mine in PLOT SETUP. 107

  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10
  • 11
  • 12
  • 13
  • 14
  • 15
  • 16
  • 17
  • 18
  • 19
  • 20
  • 21
  • 22
  • 23
  • 24
  • 25
  • 26
  • 27
  • 28
  • 29
  • 30
  • 31
  • 32
  • 33
  • 34
  • 35
  • 36
  • 37
  • 38
  • 39
  • 40
  • 41
  • 42
  • 43
  • 44
  • 45
  • 46
  • 47
  • 48
  • 49
  • 50
  • 51
  • 52
  • 53
  • 54
  • 55
  • 56
  • 57
  • 58
  • 59
  • 60
  • 61
  • 62
  • 63
  • 64
  • 65
  • 66
  • 67
  • 68
  • 69
  • 70
  • 71
  • 72
  • 73
  • 74
  • 75
  • 76
  • 77
  • 78
  • 79
  • 80
  • 81
  • 82
  • 83
  • 84
  • 85
  • 86
  • 87
  • 88
  • 89
  • 90
  • 91
  • 92
  • 93
  • 94
  • 95
  • 96
  • 97
  • 98
  • 99
  • 100
  • 101
  • 102
  • 103
  • 104
  • 105
  • 106
  • 107
  • 108
  • 109
  • 110
  • 111
  • 112
  • 113
  • 114
  • 115
  • 116
  • 117
  • 118
  • 119
  • 120
  • 121
  • 122
  • 123
  • 124
  • 125
  • 126
  • 127
  • 128
  • 129
  • 130
  • 131
  • 132
  • 133
  • 134
  • 135
  • 136
  • 137
  • 138
  • 139
  • 140
  • 141
  • 142
  • 143
  • 144
  • 145
  • 146
  • 147
  • 148
  • 149
  • 150
  • 151
  • 152
  • 153
  • 154
  • 155
  • 156
  • 157
  • 158
  • 159
  • 160
  • 161
  • 162
  • 163
  • 164
  • 165
  • 166
  • 167
  • 168
  • 169
  • 170
  • 171
  • 172
  • 173
  • 174
  • 175
  • 176
  • 177
  • 178
  • 179
  • 180
  • 181
  • 182
  • 183
  • 184
  • 185
  • 186
  • 187
  • 188
  • 189
  • 190
  • 191
  • 192
  • 193
  • 194
  • 195
  • 196
  • 197
  • 198
  • 199
  • 200
  • 201
  • 202
  • 203
  • 204
  • 205
  • 206
  • 207
  • 208
  • 209
  • 210
  • 211
  • 212
  • 213
  • 214
  • 215
  • 216
  • 217
  • 218
  • 219
  • 220
  • 221
  • 222
  • 223
  • 224
  • 225
  • 226
  • 227
  • 228
  • 229
  • 230
  • 231
  • 232
  • 233
  • 234
  • 235
  • 236
  • 237
  • 238
  • 239
  • 240
  • 241
  • 242
  • 243
  • 244
  • 245
  • 246
  • 247
  • 248
  • 249
  • 250
  • 251
  • 252
  • 253
  • 254
  • 255
  • 256
  • 257
  • 258
  • 259
  • 260
  • 261
  • 262
  • 263
  • 264
  • 265
  • 266
  • 267
  • 268
  • 269
  • 270
  • 271
  • 272
  • 273
  • 274
  • 275
  • 276
  • 277
  • 278
  • 279
  • 280
  • 281
  • 282
  • 283
  • 284
  • 285
  • 286
  • 287
  • 288
  • 289
  • 290
  • 291
  • 292
  • 293
  • 294
  • 295
  • 296
  • 297
  • 298
  • 299
  • 300
  • 301
  • 302
  • 303
  • 304
  • 305
  • 306
  • 307
  • 308
  • 309
  • 310
  • 311
  • 312
  • 313
  • 314
  • 315
  • 316
  • 317
  • 318
  • 319
  • 320
  • 321
  • 322
  • 323
  • 324
  • 325
  • 326
  • 327
  • 328
  • 329
  • 330
  • 331
  • 332
  • 333
  • 334
  • 335
  • 336
  • 337
  • 338
  • 339
  • 340
  • 341
  • 342
  • 343
  • 344
  • 345
  • 346
  • 347
  • 348
  • 349
  • 350
  • 351
  • 352
  • 353
  • 354
  • 355
  • 356
  • 357
  • 358
  • 359
  • 360
  • 361
  • 362
  • 363
  • 364
  • 365
  • 366

Multiple solutions and the initial guess
Graphing in Solve
Our first example was fairly simple because there was only one solution so it did not much matter where we
began looking for it.
When there is more than one possible answer you are required to supply an initial
estimate or guess.
The Solve aplet will then try to find a solution which is ‘near’ to the estimate.
Example 1
The volume of a cylinder is given by
V
=
2
π
rr
h
)
. Find the radius of a
cylinder which has a volume of 1 liter and a height of 10cm.
(
+
Enter the equation into
E1
as shown right.
When you are entering the
equation, ensure that you put a * sign between the
R
and the bracket.
See page 79 for more information on the reason for this.
Change to the
NUM
view and enter the known values, remembering that
1 liter=1000cm
3
. Position the highlight over
R
, enter a positive value as
your estimate, and press
to find the solution shown right of
8.57cm.
The equation is a quadratic in
R
which means two solutions
are possible. If you enter an initial estimate of -10 you will obtain the
negative solution, which is physically invalid in this case.
Example 2
If
f
x
3
(
(
)
=
x
2
x
2
5
x
+
2
find all values of x for which
f
x
)
=
1
.
Although you may have a clear picture in your mind and can provide Solve with the estimates it needs, I’ll
assume that, like me, you would find it helpful to see a graph first.
It is also possible to solve this in the Function aplet, which offers more
powerful tools. The
PLOT
view in the Solve aplet, although powerful,
can be deceptive if you don’t understand it and I sometimes find it easier
to work in the Function aplet.
In this case we will continue to work in
Solve.
In the
SYMB
view, enter the equation
Y=X^3-2X
2
-5X+2
into
E1
. In the
NUM
view, enter the known value of
Y=1
, ensure that the highlight is on
X
, making it the active variable, and then press
PLOT
. If your view
does not look like this then you may not have had the highlight on the
X
,
or your axes may not be set the same as mine in
PLOT SETUP
.
107