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$$f(x) = \frac{1}{\sqrt{1-|x|}} + \frac{1}{x^4}$$ show that $f(x) = 123$ has at least one answer. I have so far worked out the interval at which the function is continuous $(-1,0)\cup(0,1)$ The function is continuous everywhere except when $x>1$ and when $x = 0$ It is an even function since $f(-x)$ is not equal to $-f(x)$ I am really unsure how to approach this question at this point as I need two continuous values within a closed interval to state that $f(x) = 123$ exists.

Giorgos Giapitzakis
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Indigo
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  • Please use mathjax to typeset the mathematics in your post. – Lee Mosher Sep 04 '21 at 14:37
  • What happened when you made a table of values of the function, or looked at a graph of this function? Did it give you any ideas? – Lee Mosher Sep 04 '21 at 14:39
  • Near $x=-1,0,1$ you have $f(x)$ large and positive. Can you find any $x$ where $f(x)<123$? If so there is at least one solution (and it is then easy to show there are at least four solutions) – Henry Sep 04 '21 at 14:44
  • The function goes towards positive infinity as it nears 1 and 0, it also has two minimum points on either side of the y axis (symmetrical) – Indigo Sep 04 '21 at 14:51

1 Answers1

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This $f$ is admits many elementary estimates and rational values. For instance, you should be able to show that

  1. $f(\tfrac{3}{4})$ is rational, with $5 < f(\tfrac{3}{4}) < 6$;
  2. for integer $n>2$, $f(1-\tfrac{1}{n^2})$ is rational, with $n + 1 < f(1-\tfrac{1}{n^2}) < n + 2$; and
  3. for integer $n>2$, $f((\tfrac{1}{n})(2-\tfrac{1}{n}))$ is rational, with $16n^4 + 1 < f((\tfrac{1}{n})(2-\tfrac{1}{n})) < 16n^4 + 96n^3 + 2$.

With that information, pick good choices of $x$ at which to evaluate $f$ and use the intermediate value theorem with those.

K B Dave
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