finding unknown variables in f(x) = (ax + b) / (x^2 - c)

[hillaryous]

Given f(x) = (ax + b) / (x² - c) with the following properties:

i. the graph is symmetric with respect to the y-axis.
ii. the limit as x approaches 2 from the right of f(x) equals positive infinity.
iii. f'(1) = -2

determine the values of a, b, and c.

I think that c=2 because (ii.) indicates a vertical asymptote at x = 2, but I have no idea how to get the values of a or b. I plugged in 2 for c and took the derivative of f(x) and got:

f'(x) = (ax² - 4a - 2ax² - 2bx) / ((x² - 4)²)

Then I substituted 1 for x in the derivative and set it equal to -2 (iii.) and got
-18 = -5a-2b. But I don't know where to go from there, since I have one equation with two variables.

 

[stapel]

If the graph is symmetric with respect to the y-axis, so the left and right "halves" "match", then, since x² - c will be positive (above the axis) for all x-values such that x² > c, then what does this tell you about the sign on ax + b, especially as x gets large? What does this tell you about the value of "a"?

There is a vertical asymptote at x = 2. If c = 2, then the denominator is x² - 2, which has no zero at x = 2. So what must "c" actually be?

Once you have the (corrected) values of "a" and "c", take the derivative. Evaluate at x = 1, set equal to -2, and solve for the value of "b".

Eliz.

Edit: Ne'mind; complete solution posted below.

 

[hillaryous]

oops! i meant c=4.

 

[soroban]

Hello, hillaryous!

Eliz. Stapel is absolutely correct . . .

Given: \(\displaystyle \:f(x) \:= \:\frac{ax\,+\,b}{x^2\,-\,c}\) with the following properties:

1) The graph is symmetric with respect to the y-axis.

2) \(\displaystyle \lim_{x\to2^+} \:=\:+\infty\)

3) \(\displaystyle f'(1)\,=\,-2\)

Determine the values of \(\displaystyle a,\,b\) and \(\displaystyle c.\)

I think that \(\displaystyle \fbox{c\,=\,4}\) because (2) indicates a vertical asymptote at \(\displaystyle x\,=\,2\) . . . Yes!

We have: \(\displaystyle \L\:f(x) \:= \:\frac{ax\,+\,b}{x^2\,-\,4}\)

(1) Symmetry to the y-axis means that the function is "even".
. . All the \(\displaystyle x\)'s have even exponents.
. . Hence: \(\displaystyle \,\fbox{a\,=\,0}\)
We have: \(\displaystyle \:\L f(x) \:=\:\frac{b}{x^2\,-\,4}\) . . . and this satisfies (2).


We have: \(\displaystyle \:f(x) \:=\:b(x^2\,-\,4)^{-1}\)
. . .Then: \(\displaystyle \:f'(x) \:=\:-b(x^2\,-\,4)^{-2}(2x) \:=\:\frac{-2bx}{(x^2\,-\,4)^2}\)

Since \(\displaystyle (3)\;f'(1)\,=\,-2\), we have: \(\displaystyle \:f'(1) \:=\:\frac{-2b(1)}{(1^2\,-\,4)^2} \:=\:-2\)

. . and we have: \(\displaystyle \:\frac{-2b}{9}\:=\:-2\;\;\Rightarrow\;\;\fbox{b\,=\,9}\)


Therefore: \(\displaystyle \L\:\fbox{f(x)\;=\;\frac{9}{x^2\,-\,4}}\)

 

[hillaryous]

ah, yes! i got it: b=9. thank you!