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# More Algebra 2

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1.The function f(x) is invertible, but the function g(x) = kf(x) is not invertible. Find the sum of all possible values of k.

2.If f(x) and (ax+b)/(cx+d), abcd ≠ 0 and f(f(x)) = x for all x in the domain of f, what is the value of a+d?

Dec 17, 2019
edited by EpicWater  Dec 17, 2019
edited by EpicWater  Dec 17, 2019

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1. The statement of this problem is not very clear. Is k a constant? or k is another function and g(x) =(kf)(x), the product of the functions k and f? If k is a constant, then the only value of k that makes g(x)=kf(x) not invertible is k = 0, and it makes no sense to ask for the sum of all possible values of k since there is only one value, that is 0. If k is a nonzero constant, then g always has an inverse; in fact $$g^{-1}(x)=f^{-1}(\frac{x}{k})$$. For instance, we know that if $$f(x)=x^3$$, then $$f^{-1}(x)=\sqrt[3] {x}$$. The inverse of $$g(x)=kx^3$$ is just $$g^{-1}(x)= \sqrt [3]{\frac{x}{k} }$$, assuming k is not equal to zero.

If on the other hand k is a function and g(x) is the product of f and k, again the question makes no sense since there are lots of functions that when multiplied by f give a function that is not invertible. Two examples: $$g(x)=x^3\cdot \frac{1}{x}$$, where $$f(x)=x^3\ and \ k(x)= \frac{1}{x}$$; and $$g(x)=x^3 \cdot \frac{x+1}{x}$$, where $$f(x)=x^3$$ and $$k(x)= \frac{x+1}{x}$$. In both cases g is not invertible because it is not 1-1.

2. Here I will assume you mean $$f(x)= \frac{ax+b}{cx+d}$$. The answer to this question is: $$a+d=0$$. There is quite a bit of Algebra involved in showing this and it starts by setting $$f(f(x))=x$$, which means f is its own inverse. We have

$$f(f(x))=f(\frac{ax+b}{cx+d} )= \frac{a(\frac{ax+b}{cx+d})+b }{c(\frac{ax+b}{cx+d})+d } =\frac{a(ax+b)+b(cx+d)}{c(ax+b)+d(cx+d)} =\frac{x}{1}$$;

If you cross-multiply and combine terms you get,

$$(a^2+bc)x+(ab+bd)=(ca+dc)x^2+(cb+d^2)x$$.

Setting the coefficients of the polynomials on the two sides of the equation equal, we get

$$ab+bd=0$$, or $$b(a+d)=0$$,

$$ca+dc=0$$, or $$c(a+d)=0$$

$$a^2+bc=bc+d^2$$, or $$a^2-d^2=0$$.

The first two equations tell us that $$a+d=0$$, since neither b nor c could be zero (one of the assumptions is that $$abcd \neq 0$$)

Dec 17, 2019
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I'm sorry about the first question. Yes, "k" is a constant. Thank you for the help. Hope that clears things up!!

Dec 17, 2019
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I understand the second question and redid it myself. But not really the first. So what is the sum of all possible values of the constant k? Clarification would be appreciated. Thank you!!!

Dec 17, 2019