(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)
解答
(a)
At the points of intersection,
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 so
2 x ( x + 4 ) = x 2 + 1 \begin{align*}
2x(x+4)=x^2+1
\end{align*} 2 x ( x + 4 ) = x 2 + 1 Hence
x 2 + 8 x − 1 = 0 \begin{align*}
x^2+8x-1=0
\end{align*} x 2 + 8 x − 1 = 0 giving
x = − 4 ± 17 \begin{align*}
x=-4\pm\sqrt{17}
\end{align*} x = − 4 ± 17
(b)
Now
2 x x 2 + 1 − 1 x + 4 = x 2 + 8 x − 1 ( x 2 + 1 ) ( x + 4 ) = ( x + 4 − 17 ) ( x + 4 + 17 ) ( x 2 + 1 ) ( x + 4 ) \begin{align*}
&\,\frac{2x}{x^2+1}-\frac{1}{x+4}\\[4mm]
= &\,\frac{x^2+8x-1}{(x^2+1)(x+4)}\\[4mm]
= &\,\frac{(x+4-\sqrt{17})(x+4+\sqrt{17})}{(x^2+1)(x+4)}
\end{align*} = = x 2 + 1 2 x − x + 4 1 ( x 2 + 1 ) ( x + 4 ) x 2 + 8 x − 1 ( x 2 + 1 ) ( x + 4 ) ( x + 4 − 17 ) ( x + 4 + 17 ) Since x 2 + 1 > 0 x^2+1>0 x 2 + 1 > 0 x + 4 x+4 x + 4
x < − 4 − 17 or − 4 < x < − 4 + 17 \begin{align*}
x < -4-\sqrt{17}
\qquad \text{or} \qquad
-4 < x < -4+\sqrt{17}
\end{align*} x < − 4 − 17 or − 4 < x < − 4 + 17
(c)
If x < − 4 x<-4 x < − 4 1 ∣ x + 4 ∣ = − 1 x + 4 \dfrac{1}{|x+4|} = -\dfrac{1}{x+4} ∣ x + 4∣ 1 = − x + 4 1 x > − 4 x>-4 x > − 4
Therefore
x < − 4 or − 4 < x < − 4 + 17 \begin{align*}
x < -4
\qquad \text{or} \qquad
-4 < x < -4+\sqrt{17}
\end{align*} x < − 4 or − 4 < x < − 4 + 17
解答
(a)
2 x x 2 + 1 = 1 x + 4 2 x ( x + 4 ) = x 2 + 1 2 x 2 + 8 x = x 2 + 1 x 2 + 8 x − 1 = 0 \begin{align*}
\frac{2x}{x^2 + 1}
= &\,\frac{1}{x + 4}\\[4mm]
2x(x + 4)
= &\,x^2 + 1\\[4mm]
2x^2 + 8x
= &\,x^2 + 1\\[4mm]
x^2 + 8x - 1
= &\,0
\end{align*} x 2 + 1 2 x = 2 x ( x + 4 ) = 2 x 2 + 8 x = x 2 + 8 x − 1 = x + 4 1 x 2 + 1 x 2 + 1 0 Using the quadratic formula:
x = − 8 ± 8 2 − 4 ( 1 ) ( − 1 ) 2 ( 1 ) = − 8 ± 64 + 4 2 = − 8 ± 68 2 = − 8 ± 2 17 2 = − 4 ± 17 \begin{align*}
x
= &\,\frac{-8 \pm \sqrt{8^2 - 4(1)(-1)}}{2(1)}\\[4mm]
= &\,\frac{-8 \pm \sqrt{64 + 4}}{2}\\[4mm]
= &\,\frac{-8 \pm \sqrt{68}}{2}\\[4mm]
= &\,\frac{-8 \pm 2\sqrt{17}}{2}\\[4mm]
= &\,-4 \pm \sqrt{17}
\end{align*} x = = = = = 2 ( 1 ) − 8 ± 8 2 − 4 ( 1 ) ( − 1 ) 2 − 8 ± 64 + 4 2 − 8 ± 68 2 − 8 ± 2 17 − 4 ± 17
(b)
The critical values are the points of intersection x = − 4 ± 17 x = -4 \pm \sqrt{17} x = − 4 ± 17 x = − 4 x = -4 x = − 4
By considering the regions between the critical values, the solution to the inequality 2 x x 2 + 1 < 1 x + 4 \frac{2x}{x^2+1} < \frac{1}{x+4} x 2 + 1 2 x < x + 4 1
x < − 4 − 17 or − 4 < x < − 4 + 17 x < -4 - \sqrt{17} \quad \text{or} \quad -4 < x < -4 + \sqrt{17} x < − 4 − 17 or − 4 < x < − 4 + 17
(c)
Taking the absolute value ∣ x + 4 ∣ |x + 4| ∣ x + 4∣ x < − 4 x < -4 x < − 4 x x x 2 x x 2 + 1 < 1 ∣ x + 4 ∣ \frac{2x}{x^2+1} < \frac{1}{|x+4|} x 2 + 1 2 x < ∣ x + 4∣ 1 − 4 + 17 -4 + \sqrt{17} − 4 + 17 x = − 4 x = -4 x = − 4
Hence, the set of values is:
x < − 4 + 17 , x ≠ − 4 x < -4 + \sqrt{17}, \quad x \neq -4 x < − 4 + 17 , x = − 4 (Or equivalently, x ∈ ( − ∞ , − 4 ) ∪ ( − 4 , − 4 + 17 ) x \in (-\infty, -4) \cup (-4, -4 + \sqrt{17}) x ∈ ( − ∞ , − 4 ) ∪ ( − 4 , − 4 + 17 )
