I am looking to prove that $f(x) = 2xcos(2x) - (x - 2)^{2}$, on the interval $[2,3]$ and $[3,4]$ is continuous.

I know the intermediate value theorem applies, but my instructor is also asking that I prove that it is continuous. Does the existence of this intermediate value c where f(c)=0 automatically imply that the function is continuous? I dont think so.

I think I need to show continuity before applying the IVT. How do I do this?

I would appreciate any help, thanks.

  • $\begingroup$ Continuity is required to apply IVT. Using the IVT to establish continuity (a) is circular and (b) doesn't make any sense. $\endgroup$
    – Randall
    Aug 31 at 19:01
  • $\begingroup$ I agree. I'm trying to figure how to establish continuity before applying the IVT. $\endgroup$
    – lo leafs
    Aug 31 at 19:02
  • 1
    $\begingroup$ The IVT only goes in one direction, and there are counterexamples for the other direction. For instance, all derivative functions have the intermediate value property (by Darboux's theorem), but not all derivative functions are continuous. $\endgroup$ Aug 31 at 19:14

You can simply argue that the function $f_1(x)=x\cdot\cos 2x$ is continuous and the function $f_2(x)=-(x-2)^2$ is continuous, which makes their sum, $f(x)=f_1(x)+f_2(x)=x\cdot\cos 2x -(x-2)^2$ continuous as well (since sum of two continuous functions is continuous).

Note that $f_2$ is continuous because it is a polynomial in $x$ and all polynomials are continuous since $f^*(x)=x$ is continuous which implies $f_n^*(x)=x^n$ is continuous since the product of two continuous functions is continuous.

And, $f_2$ is continuous again because $f^*$ is continuous and $f_c=\cos x$ is continuous and the product of two continuous functions is continuous.

  • $\begingroup$ I think you should explain that you are also making repeated use of the fact that sums and products of continuous functions are continuous to show that your $f_1$ and $f_2$ are continuous. $\endgroup$
    – Rob Arthan
    Aug 31 at 20:50
  • $\begingroup$ @RobArthan Please check the edit. Is it okay now? $\endgroup$ Aug 31 at 21:43
  • $\begingroup$ That looks good now. $\endgroup$
    – Rob Arthan
    Aug 31 at 21:49

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