Question is a part of applications of differential calculus topic where a sign diagram of f'(x) is drawn to find when f(x) is increasing and decreasing. Answer is -1 $\leq k < 1$ but how?
Could K $\leq \frac{x^2}{1-2x}$ be a solution?
Increasing/decreasing functions: Suppose $f(x) = \frac{x+k}{x^2 + k}$ is decreasing for all x. What range of values could the constant k have?
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0Differentiate wrt $x$ and find the values of $k$ such that $f' \leq 0$ for all $x$. – 2017-02-22
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1The answer is wrong. Note that if $k=0.05$, the sequence is not decreasing for all $x$. See [here](https://www.desmos.com/calculator/g84meay560). – 2017-02-22
2 Answers
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Note that the derivative of $$\frac{x+k}{x^2+k}$$ is $$\frac{-2kx+k-x^2}{(x^2+k)^2}$$ By the quotient rule. Since $(x^2+k)^2 \ge 0$, if the function is decreasing we have $$-x^2-2kx+k \le 0 \iff x^2+2kx-k \ge 0$$ For all $x$. Since this polynomial is always positive, the discriminant must not be positive: $$k^2+k \le 0$$ So we have $-1 \le k \le 0$.
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0Mad respect, thanks a lot for the solution and finding an error in the textbook :) – 2017-02-22
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0@Zephirum I would appreciate it if you were to accept this answer by clicking on the check mark below the upvote button :) Thanks. – 2017-02-22
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We have $$f'(x) = \frac {(x^2+k)(1)-(x+k)(2x)}{(x^2+k)^2} = \frac {-x^2-2xk+k}{(x^2+k)^2} = \frac {(k^2+k)-(x+k)^2}{(x^2+k)^2} $$
Now, if $f (x) $ is decreasing, then $f'(x) \leq 0$ for all $x $. Hope you can take it from here.