题目
Problem
In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
The function f is defined by
f(x)=5(x2−2)4x+9x≥−49
(a) Show that
f′(x)=4x+9k(5x2+9x−2)
where k is an integer to be found.
(4)
(b) Hence, find the values of x for which f′(x)=0
(1)
Figure 3 shows a sketch of the curve C with equation y=f(x).
The curve has a local maximum at the point P
(c) Find the exact coordinates of P
(2)
The function g is defined by
g(x)=2f(x)+4−49≤x≤0
(d) Find the range of g
(3)
题目中文翻译
图 3 给出了函数图像的一部分。
本题中你必须写出解题过程的所有步骤。
不接受完全依赖计算器技术的解法。
函数 f 定义为
f(x)=5(x2−2)4x+9x≥−49
(a) 证明
f′(x)=4x+9k(5x2+9x−2)
其中 k 是需要求出的整数。
(b) 由此求使 f′(x)=0 的 x 值。
图 3 给出了曲线 C（方程为 y=f(x)）的示意图。
曲线在点 P 处有一个局部极大值。
(c) 求点 P 的精确坐标。
函数 g 定义为
g(x)=2f(x)+4−49≤x≤0
(d) 求 g 的值域。
解答
(a)
We have
f(x)=5(x2−2)(4x+9)1/2
Using the product rule,
f′(x)=10x(4x+9)1/2+5(x2−2)⋅21(4x+9)−1/2⋅4
So
f′(x)=10x(4x+9)1/2+10(x2−2)(4x+9)−1/2
Put over the common denominator 4x+9:
f′(x)=4x+910x(4x+9)+10(x2−2)=4x+940x2+90x+10x2−20=4x+950x2+90x−20=4x+910(5x2+9x−2)
Therefore
k=10
(b)
From part (a),
f′(x)=0
when
5x2+9x−2=0
Factorise:
(5x−1)(x+2)=0
So
x=51
or
x=−2
Therefore
x=−2, 51
(c)
The local maximum is at the smaller stationary value,
x=−2
Using
f(x)=5(x2−2)4x+9
we get
f(−2)=5(4−2)1
So
f(−2)=10
Therefore
P=(−2,10)
(d)
On
−49≤x≤0
the maximum value of f is
10
from part (c).
Also,
f(0)=5(02−2)9=−30
So the range of f on this interval is
−30≤f(x)≤10
Since
g(x)=2f(x)+4
the lower bound is
2(−30)+4=−56
and the upper bound is
2(10)+4=24
Therefore the range of g is
−56≤g(x)≤24 