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)

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:

    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);

    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);

    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。

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