题目
Problem
The curve C C C y = f ( x ) y=f(x) y = f ( x )
f ( x ) = x 3 4 x + 7 x ≥ − 7 4 f(x)=x^3\sqrt{4x+7}\qquad x\geq -\frac{7}{4} f ( x ) = x 3 4 x + 7 x ≥ − 4 7 (a) Show that
f ′ ( x ) = k x 2 ( 2 x + 3 ) 4 x + 7 f'(x)=\frac{kx^2(2x+3)}{\sqrt{4x+7}} f ′ ( x ) = 4 x + 7 k x 2 ( 2 x + 3 ) where k k k
(4)
The point P P P C C C
(b) Find the coordinates of P P P
(2)
(c) Hence find the range of the function g g g
g ( x ) = − 4 f ( x ) x ≥ − 7 4 g(x)=-4f(x)\qquad x\geq -\frac{7}{4} g ( x ) = − 4 f ( x ) x ≥ − 4 7 (2)
The point Q Q Q ( 1 2 , 3 8 ) \left(\dfrac12,\dfrac38\right) ( 2 1 , 8 3 ) C C C
(d) Find the coordinates of the point to which Q Q Q C C C
y = 40 f ( x − 3 2 ) − 8 y=40f\left(x-\frac32\right)-8 y = 40 f ( x − 2 3 ) − 8 (2)
题目中文翻译
曲线 C C C y = f ( x ) y=f(x) y = f ( x )
f ( x ) = x 3 4 x + 7 x ≥ − 7 4 f(x)=x^3\sqrt{4x+7}\qquad x\geq -\frac{7}{4} f ( x ) = x 3 4 x + 7 x ≥ − 4 7 (a) 证明
f ′ ( x ) = k x 2 ( 2 x + 3 ) 4 x + 7 f'(x)=\frac{kx^2(2x+3)}{\sqrt{4x+7}} f ′ ( x ) = 4 x + 7 k x 2 ( 2 x + 3 ) 其中 k k k
图 3 所示的点 P P P C C C
(b) 求点 P P P
(c) 由此求函数 g g g
g ( x ) = − 4 f ( x ) x ≥ − 7 4 g(x)=-4f(x)\qquad x\geq -\frac{7}{4} g ( x ) = − 4 f ( x ) x ≥ − 4 7 坐标为 ( 1 2 , 3 8 ) \left(\dfrac12,\dfrac38\right) ( 2 1 , 8 3 ) Q Q Q C C C
(d) 当 C C C
y = 40 f ( x − 3 2 ) − 8 y=40f\left(x-\frac32\right)-8 y = 40 f ( x − 2 3 ) − 8 所表示的曲线时，求点 Q Q Q
解答
(a)
解法一
思路
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把根式写成 ( 4 x + 7 ) 1 / 2 (4x+7)^{1/2} ( 4 x + 7 ) 1/2 x 3 ( 4 x + 7 ) 1 / 2 x^3(4x+7)^{1/2} x 3 ( 4 x + 7 ) 1/2
答题过程
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Write
f ( x ) = x 3 ( 4 x + 7 ) 1 2 . f(x)=x^3(4x+7)^{\frac12}. f ( x ) = x 3 ( 4 x + 7 ) 2 1 . Using the product rule,
f ′ ( x ) = 3 x 2 ( 4 x + 7 ) 1 2 + 2 x 3 ( 4 x + 7 ) − 1 2 = 3 x 2 ( 4 x + 7 ) + 2 x 3 4 x + 7 = 7 x 2 ( 2 x + 3 ) 4 x + 7 . \begin{align*}
f'(x)
=&\,3x^2(4x+7)^{\frac12} \\
&\,+2x^3(4x+7)^{-\frac12} \\[2mm]
=&\,\frac{3x^2(4x+7)+2x^3}
{\sqrt{4x+7}} \\[2mm]
=&\,\frac{7x^2(2x+3)}
{\sqrt{4x+7}}.
\end{align*} f ′ ( x ) = = = 3 x 2 ( 4 x + 7 ) 2 1 + 2 x 3 ( 4 x + 7 ) − 2 1 4 x + 7 3 x 2 ( 4 x + 7 ) + 2 x 3 4 x + 7 7 x 2 ( 2 x + 3 ) . Therefore,
k = 7 . \boxed{k=7}. k = 7 .
(b)
解法一
思路
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由 (a) 的导数求驻点。分母在定义域内部为正，而 x 2 ≥ 0 x^2\geq0 x 2 ≥ 0 2 x + 3 2x+3 2 x + 3 x = − 3 2 x=-\frac32 x = − 2 3
答题过程
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For an interior stationary point,
7 x 2 ( 2 x + 3 ) 4 x + 7 = 0. \frac{7x^2(2x+3)}{\sqrt{4x+7}}=0. 4 x + 7 7 x 2 ( 2 x + 3 ) = 0. Thus,
x = 0 x=0 x = 0 or
x = − 3 2 . x=-\frac32. x = − 2 3 . For − 7 4 < x < − 3 2 -\frac74<x<-\frac32 − 4 7 < x < − 2 3 f ′ ( x ) < 0 f'(x)<0 f ′ ( x ) < 0 x > − 3 2 x>-\frac32 x > − 2 3 f ′ ( x ) ≥ 0 f'(x)\geq0 f ′ ( x ) ≥ 0 x = 0 x=0 x = 0 x = − 3 2 x=-\frac32 x = − 2 3
Its y y y
f ( − 3 2 ) = ( − 3 2 ) 3 4 ( − 3 2 ) + 7 = − 27 8 . \begin{align*}
f\bigg(-\frac32\bigg)
=&\,\bigg(-\frac32\bigg)^3
\sqrt{4\big(-\frac32\big)+7} \\[2mm]
=&\,-\frac{27}{8}.
\end{align*} f ( − 2 3 ) = = ( − 2 3 ) 3 4 ( − 2 3 ) + 7 − 8 27 . Therefore,
P ( − 3 2 , − 27 8 ) . \boxed{P\bigg(-\frac32,-\frac{27}{8}\bigg)}. P ( − 2 3 , − 8 27 ) .
(c)
解法一
思路
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本小题必须承接 (b)。由于 f f f − 27 8 -\frac{27}{8} − 8 27 f ( x ) ≥ − 27 8 f(x)\geq-\frac{27}{8} f ( x ) ≥ − 8 27 − 4 -4 − 4 g g g f f f g g g
答题过程
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From part (b),
f ( x ) ≥ − 27 8 . f(x)\geq-\frac{27}{8}. f ( x ) ≥ − 8 27 . Multiplying by − 4 -4 − 4
− 4 f ( x ) ≤ 27 2 . -4f(x)\leq\frac{27}{2}. − 4 f ( x ) ≤ 2 27 . Since f ( x ) → ∞ f(x)\to\infty f ( x ) → ∞ x → ∞ x\to\infty x → ∞ g ( x ) = − 4 f ( x ) → − ∞ g(x)=-4f(x)\to-\infty g ( x ) = − 4 f ( x ) → − ∞ g g g
( − ∞ , 27 2 ] . \boxed{\big(-\infty,\frac{27}{2}\big]}. ( − ∞ , 2 27 ] .
(d)
解法一
思路
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f ( x − 3 2 ) f(x-\frac32) f ( x − 2 3 ) 3 2 \frac32 2 3 3 2 \frac32 2 3 40 40 40 40 40 40 − 8 -8 − 8 8 8 8
答题过程
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The transformation
y = 40 f ( x − 3 2 ) − 8 y=40f\bigg(x-\frac32\bigg)-8 y = 40 f ( x − 2 3 ) − 8 maps a point ( x , y ) (x,y) ( x , y ) C C C
( x + 3 2 , 40 y − 8 ) . \bigg(x+\frac32,\,40y-8\bigg). ( x + 2 3 , 40 y − 8 ) . Therefore,
( 1 2 , 3 8 ) ⟼ ( 1 2 + 3 2 , 40 ( 3 8 ) − 8 ) = ( 2 , 7 ) . \begin{align*}
\bigg(\frac12,\frac38\bigg)
&\longmapsto
\bigg(\frac12+\frac32,\,
40\big(\frac38\big)-8\bigg) \\[2mm]
&=\boxed{(2,7)}.
\end{align*} ( 2 1 , 8 3 ) ⟼ ( 2 1 + 2 3 , 40 ( 8 3 ) − 8 ) = ( 2 , 7 ) . 
