Question

请帮忙解一下微积分的题

let f be the function that is given by f(x)=(ax+b)/(x^2-c) and that has the following properties:
1) the graph of f is symmetric with respect to the y-axis
2)lim x-> 2+ f(x)=+00（正无穷）
3）f(1)=-2
a)determine the values of a,b,c
b)write an equation for each vertical ad each horizontal asymtptote of the graph of f
c) sketch the graph of f in the xy-plane provided below(这个可不答)

Answer 1

您好，【科技英语强人团】树的儿子为您解答如下：

a) 分析：要求a，b，c的值，已知有3个条件，我们一一来看：

1. The graph of f is symmetric with respect to the y-axis. 即f（x）的图象关于y轴对称，也就是说f（x）是个偶函数（even function)。那么f（-x）=f（x）。将x和-x代入函数表达式f(x)=(ax+b)/(x^2-c)，建立等式，化简，得：-a=a，则a=0。于是f（x）=b/(x^2-c) （1）；

2. lim x-> 2+  f(x)=+00（正无穷），即当x趋近于2时，函数值趋向于正无穷。由于分子部分是一个定值b，代入x=2不会出现0/0这样的indeterminant form，那么可能使得函数值趋向于正无穷的只有可能是分母为0，即当x=2时，x^2-c=0，解出c=4。于是f（x）=b/(x^2-4）（2）；

3. f(1)=-2，直接代入（2）解出b=6。

综上，a=0,b=6,c=4, f(x)=6/(x^2-4).

英文解答：As the graph of f is symmetric with respect to the y-axis, f(x)=f(-x). Plug in x and -x to f(x), we get the equation (ax+b)/(x^2-c)=(-ax+b)/(x^2-c). Simplify it, and we get -a=a. Solve for a: a=0;

Now the numerator part of the function collapes to a single constant b. As lim x-> 2+  f(x) equals positive infinity, the denominator has to be 0 when we plug in x=2. Set up equation, and solve for c--we get c=4;

Finally we plug in 1 for x and the function should be equal to -2. Solve for b and we get b=6.

Thus, a=0,b=6,c=4, and f(x)=6/(x^2-4).

 

b) Write an equation for each vertical ad each horizontal asymtptote of the graph of f. 让你写出f（x）所有水平及垂直渐近线的方程。

分析：用u代替x^2, f（x）就是关于u的反比例函数，此时f(u)=6/(u-4)，且u的范围是u>=0. 根据函数的平移特性可知，f(u)=6/(u-4)就是f(x)=6/u向右平移4个单位，其垂直渐近线就是u=4，即x^2=4，x=+/-2；由于没有进行垂直平移，f(u)=6/(u-4)的水平渐近线就是f(u)=6/u的水平渐近线，即y=0.

 

英文解答：Substitute x^2 with u. Since we know that f(u)=6/(u-4) has vertical asymptote u=4, f(x)=6/(x^2-4) should have two vertical symptotes such that x^2=4--x=+/-2.

Since f(u)=6/(u-4) has identical horizontal asymptote as f(u)=6/u does, f(x) has horizontal asymptote y=0.

Thus, the equations for vertical and horizontal asymptotes of the graph of f are x=-2, x-2 and y=0.

 c)注意是偶函数就可以了。自己求一阶导数和二阶导数分别找出monotonity 和 concavity，可以画出图象。