题目
Problem
2.
(a) Use algebra to determine the exact x x x
y = 2 x x 2 + 1 and y = 1 x + 4 \begin{align*}
y=\frac{2x}{x^2+1}
\qquad \text{and} \qquad
y=\frac{1}{x+4}
\end{align*} y = x 2 + 1 2 x and y = x + 4 1 (3)
Hence
(b) determine the values of x x x
2 x x 2 + 1 < 1 x + 4 \begin{align*}
\frac{2x}{x^2+1}<\frac{1}{x+4}
\end{align*} x 2 + 1 2 x < x + 4 1 (2)
(c) state the values of x x x
2 x x 2 + 1 < 1 ∣ x + 4 ∣ \begin{align*}
\frac{2x}{x^2+1}<\frac{1}{|x+4|}
\end{align*} x 2 + 1 2 x < ∣ x + 4∣ 1 (2)
题目中文翻译
(a) 使用代数方法，求以下两条曲线交点的精确 x x x
y = 2 x x 2 + 1 , y = 1 x + 4 . \begin{align*}
y=&\,\frac{2x}{x^2+1},\\[3mm]
y=&\,\frac1{x+4}.
\end{align*} y = y = x 2 + 1 2 x , x + 4 1 . (3)
由此，
(b) 求满足下列不等式的 x x x
2 x x 2 + 1 < 1 x + 4 . \begin{align*}
\frac{2x}{x^2+1}<\frac1{x+4}.
\end{align*} x 2 + 1 2 x < x + 4 1 . (2)
(c) 写出满足下列不等式的 x x x
2 x x 2 + 1 < 1 ∣ x + 4 ∣ . \begin{align*}
\frac{2x}{x^2+1}<\frac1{|x+4|}.
\end{align*} x 2 + 1 2 x < ∣ x + 4∣ 1 . (2)
解答
(a)
解法一
思路
展开
交点的 x x x y y y x 2 + 1 x^2+1 x 2 + 1 x + 4 x+4 x + 4 0 0 0
答题过程
展开
At the points of intersection,
2 x x 2 + 1 = 1 x + 4 \begin{align*}
\frac{2x}{x^2+1}
=&\,
\frac1{x+4}
\end{align*} x 2 + 1 2 x = x + 4 1 Cross-multiply:
2 x ( x + 4 ) = x 2 + 1 2 x 2 + 8 x = x 2 + 1 x 2 + 8 x − 1 = 0 \begin{align*}
2x(x+4)
=&\,
x^2+1\\[2mm]
2x^2+8x
=&\,
x^2+1\\[2mm]
x^2+8x-1
=&\,
0
\end{align*} 2 x ( x + 4 ) = 2 x 2 + 8 x = x 2 + 8 x − 1 = x 2 + 1 x 2 + 1 0 Using the quadratic formula,
x = − 8 ± 8 2 − 4 ( 1 ) ( − 1 ) 2 = − 8 ± 68 2 = − 4 ± 17 \begin{align*}
x
=&\,
\frac{-8\pm\sqrt{8^2-4(1)(-1)}}{2}\\[3mm]
=&\,
\frac{-8\pm\sqrt{68}}{2}\\[3mm]
=&\,
-4\pm\sqrt{17}
\end{align*} x = = = 2 − 8 ± 8 2 − 4 ( 1 ) ( − 1 ) 2 − 8 ± 68 − 4 ± 17 Therefore the exact x x x
x = − 4 − 17 , x = − 4 + 17 \begin{align*}
\boxed{x=-4-\sqrt{17},\quad x=-4+\sqrt{17}}
\end{align*} x = − 4 − 17 , x = − 4 + 17
解法二
思路
展开
将两式移到等号同一边，通过通分化为单一分式，即 2 x x 2 + 1 − 1 x + 4 = 0 \dfrac{2x}{x^2+1} - \dfrac{1}{x+4} = 0 x 2 + 1 2 x − x + 4 1 = 0 0 0 0 x x x
答题过程
展开
At the points of intersection, write the equation by moving all terms to one side:
2 x x 2 + 1 − 1 x + 4 = 0 \begin{align*}
\frac{2x}{x^2+1} - \frac{1}{x+4}
=&\,
0
\end{align*} x 2 + 1 2 x − x + 4 1 = 0 Combine the fractions over a common denominator:
2 x ( x + 4 ) − ( x 2 + 1 ) ( x 2 + 1 ) ( x + 4 ) = 0 2 x 2 + 8 x − x 2 − 1 ( x 2 + 1 ) ( x + 4 ) = 0 x 2 + 8 x − 1 ( x 2 + 1 ) ( x + 4 ) = 0 \begin{align*}
\frac{2x(x+4) - (x^2+1)}{(x^2+1)(x+4)}
=&\,
0\\[4mm]
\frac{2x^2+8x-x^2-1}{(x^2+1)(x+4)}
=&\,
0\\[4mm]
\frac{x^2+8x-1}{(x^2+1)(x+4)}
=&\,
0
\end{align*} ( x 2 + 1 ) ( x + 4 ) 2 x ( x + 4 ) − ( x 2 + 1 ) = ( x 2 + 1 ) ( x + 4 ) 2 x 2 + 8 x − x 2 − 1 = ( x 2 + 1 ) ( x + 4 ) x 2 + 8 x − 1 = 0 0 0 Since the denominator cannot be zero at the points of intersection, set the numerator to zero:
x 2 + 8 x − 1 = 0 \begin{align*}
x^2+8x-1
=&\,
0
\end{align*} x 2 + 8 x − 1 = 0 Solve using the quadratic formula:
x = − 8 ± 8 2 − 4 ( 1 ) ( − 1 ) 2 = − 8 ± 68 2 = − 4 ± 17 \begin{align*}
x
=&\,
\frac{-8\pm\sqrt{8^2-4(1)(-1)}}{2}\\[4mm]
=&\,
\frac{-8\pm\sqrt{68}}{2}\\[4mm]
=&\,
-4\pm\sqrt{17}
\end{align*} x = = = 2 − 8 ± 8 2 − 4 ( 1 ) ( − 1 ) 2 − 8 ± 68 − 4 ± 17 Therefore the exact x x x
x = − 4 − 17 , x = − 4 + 17 \begin{align*}
\boxed{x=-4-\sqrt{17},\quad x=-4+\sqrt{17}}
\end{align*} x = − 4 − 17 , x = − 4 + 17
(b)
解法一
思路
展开
用 (a) 的两个交点，再加上 x = − 4 x=-4 x = − 4 x 2 + 1 > 0 x^2+1>0 x 2 + 1 > 0 x + 4 x+4 x + 4
答题过程
展开
Move everything to one side:
2 x x 2 + 1 − 1 x + 4 < 0 2 x ( x + 4 ) − ( x 2 + 1 ) ( x 2 + 1 ) ( x + 4 ) < 0 x 2 + 8 x − 1 ( x 2 + 1 ) ( x + 4 ) < 0 \begin{align*}
\frac{2x}{x^2+1}-\frac1{x+4}
<&
0\\[3mm]
\frac{2x(x+4)-(x^2+1)}
{(x^2+1)(x+4)}
<&
0\\[3mm]
\frac{x^2+8x-1}
{(x^2+1)(x+4)}
<&
0
\end{align*} x 2 + 1 2 x − x + 4 1 < ( x 2 + 1 ) ( x + 4 ) 2 x ( x + 4 ) − ( x 2 + 1 ) < ( x 2 + 1 ) ( x + 4 ) x 2 + 8 x − 1 < 0 0 0 Since
x 2 + 8 x − 1 = ( x + 4 − 17 ) ( x + 4 + 17 ) , \begin{align*}
x^2+8x-1
=&\,
\bigg(x+4-\sqrt{17}\bigg)
\bigg(x+4+\sqrt{17}\bigg),
\end{align*} x 2 + 8 x − 1 = ( x + 4 − 17 ) ( x + 4 + 17 ) , the critical values are
− 4 − 17 , − 4 , − 4 + 17 \begin{align*}
-4-\sqrt{17},\qquad -4,\qquad -4+\sqrt{17}
\end{align*} − 4 − 17 , − 4 , − 4 + 17 Let a = − 4 − 17 a=-4-\sqrt{17} a = − 4 − 17 b = − 4 + 17 b=-4+\sqrt{17} b = − 4 + 17 x 2 + 1 > 0 x^2+1>0 x 2 + 1 > 0 x x x
Interval x 2 + 8 x − 1 x^2+8x-1 x 2 + 8 x − 1 ( x 2 + 1 ) ( x + 4 ) (x^2+1)(x+4) ( x 2 + 1 ) ( x + 4 ) Quotient x < a x<a x < a + + + − - − − - − a < x < − 4 a<x<-4 a < x < − 4 − - − − - − + + + − 4 < x < b -4<x<b − 4 < x < b − - − + + + − - − x > b x>b x > b + + + + + + + + +
The numerator vanishes at a a a b b b x = − 4 x=-4 x = − 4
x < − 4 − 17 or − 4 < x < − 4 + 17 \begin{align*}
\boxed{
x<-4-\sqrt{17}
\quad \text{or} \quad
-4<x<-4+\sqrt{17}
}
\end{align*} x < − 4 − 17 or − 4 < x < − 4 + 17
(c)
解法一
思路
展开
绝对值会把 1 x + 4 \frac1{x+4} x + 4 1 x > − 4 x>-4 x > − 4 x < − 4 x<-4 x < − 4 − 1 x + 4 -\frac1{x+4} − x + 4 1 x < 0 x<0 x < 0
答题过程
展开
For x > − 4 x>-4 x > − 4
∣ x + 4 ∣ = x + 4 \begin{align*}
|x+4|=x+4
\end{align*} ∣ x + 4∣ = x + 4 so the inequality is the same as in part (b). This gives
− 4 < x < − 4 + 17 \begin{align*}
-4<x<-4+\sqrt{17}
\end{align*} − 4 < x < − 4 + 17 For x < − 4 x<-4 x < − 4
∣ x + 4 ∣ = − ( x + 4 ) \begin{align*}
|x+4|=-(x+4)
\end{align*} ∣ x + 4∣ = − ( x + 4 ) so
1 ∣ x + 4 ∣ > 0 \begin{align*}
\frac1{|x+4|}>0
\end{align*} ∣ x + 4∣ 1 > 0 But for x < − 4 x<-4 x < − 4
2 x x 2 + 1 < 0 \begin{align*}
\frac{2x}{x^2+1}<0
\end{align*} x 2 + 1 2 x < 0 Therefore the inequality is true for all x < − 4 x<-4 x < − 4 x = − 4 x=-4 x = − 4
x < − 4 or − 4 < x < − 4 + 17 \begin{align*}
\boxed{
x<-4
\quad \text{or} \quad
-4<x<-4+\sqrt{17}
}
\end{align*} x < − 4 or − 4 < x < − 4 + 17 