Mathematics
Senior High
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連立方程式が出来ません
3つとか4つの連立方程式の解き方教えてください💦
28
連立
x=1で極大値 6, x=2で極小値5をとるような3次関数f(x) を求めよ。
406
fra) = ax ³ + b ² + cx td (a 40) 245.
f'(x) = 3ax² +26* *c
*-12"#66.44.
f(1) = 6 J/(1) =0
a + b + cid = 6
3A +26+0 = 0
x=2で極小値5より
$(2) = 5 5' (₂) 20
8a+ 4b +2ord +5
120 +4670
20+20+20+20 012
3a+2b1c20
- arcted = [2
12
Answers
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