4.11: Kazi ya Ufafanuzi wa Kipande
- Page ID
- 164612
Kazi zilizoelezwa kwa kipande ni kazi ambazo hufafanuliwa kwa kutumia milinganyo tofauti kwa sehemu tofauti za kikoa.
Tathmini zifuatazo piecewise-defined kazi kwa maadili fulani ya\(x\), na graph kazi:
\(f(x) = \left\{\begin{array}{cc}−2x + 1 & −1 \leq x < 0 \\ x^2 + 2 &0 \leq x \leq 2\end{array} \right.\)
Suluhisho
Ili kuchora kazi hii, fanya meza ya ufumbuzi:
Jedwali la Solutions kwa\(f(x) = −2x + 1 \) Domain\(−1 \leq x < 0\) |
|
\(x\) | \(f(x)\) |
-1 | 3 |
0 | 1 (mduara wazi hapa, 0 si katika uwanja) |
Jedwali la Solutions kwa\(f(x) = x^2 + 2\) Domain\(0 \leq x \leq 2\) |
|
\(x\) | \(f(x)\) |
0 | 2 |
1 | 3 |
2 | 6 |
Tathmini zifuatazo piecewise-defined kazi kwa maadili fulani ya\(x\), na graph kazi:
\(f(x) = \left\{\begin{array}{cc} −x + 1 &x \leq −1 \\ 2 & −1 < x \leq 1 \\ −x + 3 &x > 1 \end{array}\right.\)
Suluhisho
Ili kuchora kazi hii, tena ufanye meza ya ufumbuzi:
Jedwali la Solutions kwa\(f(x) = −x + 1\) Domain\(x \leq −1\) |
|
\(x\) | \(f(x)\) |
-3 | 4 |
-2 | 3 |
-1 | 2 (mduara uliofungwa hapa, -1 iko kwenye kikoa) |
Jedwali la Solutions kwa\(f(x) = 2\) Domain\(−1 < x \leq 1\) |
|
\(x\) | \(f(x)\) |
-1 | 2 (mduara wazi kujazwa na kazi ya awali, -1 si katika uwanja) |
0 | 2 |
1 | 2 (mduara uliofungwa hapa, 1 iko katika uwanja) |
Jedwali la Solutions kwa\(f(x) = −x + 3\) Domain\(x > 1\) |
|
\(x\) | \(f(x)\) |
1 | 2 (mduara wazi kujazwa na kazi ya awali, 1 si katika uwanja) |
2 | 1 |
3 | 0 |
Tathmini kazi zifuatazo zilizoelezwa kwa kipande kwa maadili yaliyotolewa ya x, na graph kazi:.
- \ (f (x) =\ kushoto\ {\ kuanza {safu} {cc}
x & x<0\\
2 x+1 &x\ geq 0
\ mwisho {safu}\ haki.\) - \(g(x) = \left\{\begin{array}{cc} 4 − x& x < 2\\ 2x − 2 &x \geq 2 \end{array} \right.\)
- \(h(x) = \left\{\begin{array}{cc} −x − 1 & x < −1 \\ 0& −1 \leq x \leq 1 \\ x + 1 & x > 1 \end{array} \right.\)
- \(g(x) = \left\{\begin{array}{cc} 6 & −8 \leq x < −4 \\ 3 &−4 \leq x \leq 5 \end{array}\right.\)
- \(f(x) = \left\{\begin{array}{cc} −x + 1 & −1 \leq x < 1 \\ \sqrt{x − 1 } &1 \leq x \leq 5\end{array}\right.\)