题目
Problem
The functions f f f g g g
f ( x ) = 4 x + 6 x − 5 x ∈ R , x ≠ 5 f(x)=\frac{4x+6}{x-5}\qquad x\in\mathbb{R},\ x\ne 5 f ( x ) = x − 5 4 x + 6 x ∈ R , x = 5 g ( x ) = 5 − 2 x 2 x ∈ R , x ≤ 0 g(x)=5-2x^2\qquad x\in\mathbb{R},\ x\le 0 g ( x ) = 5 − 2 x 2 x ∈ R , x ≤ 0 (a) Solve the equation
f g ( x ) = 3 fg(x)=3 f g ( x ) = 3 (4)
(b) Find f − 1 f^{-1} f − 1
(3)
(c) Sketch and label, on the same axes, the curve with equation y = g ( x ) y=g(x) y = g ( x ) y = g − 1 ( x ) y=g^{-1}(x) y = g − 1 ( x )
(3)
题目中文翻译
函数 f f f g g g
f ( x ) = 4 x + 6 x − 5 x ∈ R , x ≠ 5 f(x)=\frac{4x+6}{x-5}\qquad x\in\mathbb{R},\ x\ne 5 f ( x ) = x − 5 4 x + 6 x ∈ R , x = 5 g ( x ) = 5 − 2 x 2 x ∈ R , x ≤ 0 g(x)=5-2x^2\qquad x\in\mathbb{R},\ x\le 0 g ( x ) = 5 − 2 x 2 x ∈ R , x ≤ 0 (a) 解方程
f g ( x ) = 3 fg(x)=3 f g ( x ) = 3 (b) 求 f − 1 f^{-1} f − 1
(c) 在同一坐标轴上画出并标注曲线 y = g ( x ) y=g(x) y = g ( x ) y = g − 1 ( x ) y=g^{-1}(x) y = g − 1 ( x )
解答
(a)
We need to solve
f g ( x ) = 3 fg(x)=3 f g ( x ) = 3 This means
f ( g ( x ) ) = 3 f(g(x))=3 f ( g ( x )) = 3 Since
g ( x ) = 5 − 2 x 2 g(x)=5-2x^2 g ( x ) = 5 − 2 x 2 we have
f ( g ( x ) ) = 4 ( 5 − 2 x 2 ) + 6 ( 5 − 2 x 2 ) − 5 f(g(x))=\frac{4(5-2x^2)+6}{(5-2x^2)-5} f ( g ( x )) = ( 5 − 2 x 2 ) − 5 4 ( 5 − 2 x 2 ) + 6 Simplify:
f ( g ( x ) ) = 20 − 8 x 2 + 6 − 2 x 2 = 26 − 8 x 2 − 2 x 2 \begin{aligned}
f(g(x))
&=\frac{20-8x^2+6}{-2x^2}\\
&=\frac{26-8x^2}{-2x^2}
\end{aligned} f ( g ( x )) = − 2 x 2 20 − 8 x 2 + 6 = − 2 x 2 26 − 8 x 2 Set this equal to 3 3 3
26 − 8 x 2 − 2 x 2 = 3 \frac{26-8x^2}{-2x^2}=3 − 2 x 2 26 − 8 x 2 = 3 Multiply by − 2 x 2 -2x^2 − 2 x 2
26 − 8 x 2 = − 6 x 2 26-8x^2=-6x^2 26 − 8 x 2 = − 6 x 2 So
26 = 2 x 2 26=2x^2 26 = 2 x 2 Therefore
x 2 = 13 x^2=13 x 2 = 13 Since the domain of g g g
x ≤ 0 x\leq 0 x ≤ 0 we take the negative root.
Hence
x = − 13 \boxed{x=-\sqrt{13}} x = − 13 (b)
Let
y = f ( x ) y=f(x) y = f ( x ) Then
y = 4 x + 6 x − 5 y=\frac{4x+6}{x-5} y = x − 5 4 x + 6 Rearrange:
y ( x − 5 ) = 4 x + 6 y(x-5)=4x+6 y ( x − 5 ) = 4 x + 6 So
y x − 5 y = 4 x + 6 yx-5y=4x+6 y x − 5 y = 4 x + 6 Bring the x x x
y x − 4 x = 5 y + 6 yx-4x=5y+6 y x − 4 x = 5 y + 6 Factorise:
x ( y − 4 ) = 5 y + 6 x(y-4)=5y+6 x ( y − 4 ) = 5 y + 6 Therefore
x = 5 y + 6 y − 4 x=\frac{5y+6}{y-4} x = y − 4 5 y + 6 Replace y y y x x x
f − 1 ( x ) = 5 x + 6 x − 4 \boxed{f^{-1}(x)=\frac{5x+6}{x-4}} f − 1 ( x ) = x − 4 5 x + 6 The domain of f − 1 f^{-1} f − 1
x ∈ R , x ≠ 4 \boxed{x\in\mathbb{R},\ x\ne 4} x ∈ R , x = 4 (c)
For
g ( x ) = 5 − 2 x 2 , x ≤ 0 g(x)=5-2x^2,\qquad x\leq 0 g ( x ) = 5 − 2 x 2 , x ≤ 0 the graph is the left half of a downward-opening parabola.
Its y y y
( 0 , 5 ) (0,5) ( 0 , 5 ) For the x x x
5 − 2 x 2 = 0 5-2x^2=0 5 − 2 x 2 = 0 so
x 2 = 5 2 x^2=\frac52 x 2 = 2 5 Since x ≤ 0 x\leq 0 x ≤ 0
x = − 5 2 x=-\sqrt{\frac52} x = − 2 5 Thus the x x x
( − 5 2 , 0 ) \left(-\sqrt{\frac52},0\right) ( − 2 5 , 0 ) The graph of y = g − 1 ( x ) y=g^{-1}(x) y = g − 1 ( x ) y = g ( x ) y=g(x) y = g ( x ) y = x y=x y = x
Therefore its intercepts are
( 5 , 0 ) (5,0) ( 5 , 0 ) and
( 0 , − 5 2 ) \left(0,-\sqrt{\frac52}\right) ( 0 , − 2 5 )
