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Show $tan(\pi x) − x − 6 = 0$ has a root between $0$ and $1$

for this question when I plug in $0$ or $1$ as values for $x$, $f(x)$ gives the values of $-6$ and $-7$, respectively so because it does not appear to cross the x-axis, I cannot say there is a root between $0$ and $1$ using Intermediate Value Theorem.

Is there another method I should be using?

Mathew Mahindaratne
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number8
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    Hint: the tan function has asymptotes. Are there any for this function between $0$ and $1$? – Minus One-Twelfth Mar 03 '19 at 14:07
  • The IVT does not apply here because $\tan(\pi x)$ is not defined at $x=\frac{1}{2}$ (and therefore not continous which is a crucial condition for IVT) – Peter Mar 03 '19 at 14:09
  • But we can see that at $x=0.5$ (infinity is positive) it is positive and for $x=0$ it is negative. – Abhinav Mar 03 '19 at 14:12
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    @Abhinav The sign in this case depends on which side of $0.5$ we are (if we are very near to $0.5$). So, to say that at $x=0.5$ , the value is positive, is not correct (even if we consider $\infty$ to be a "number"). Correct is : The limit for $x\rightarrow 0.5$ from the left is $+\infty$ and $f(0)<0$ – Peter Mar 03 '19 at 14:59

4 Answers4

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$\tan(x)$ can be written in terms of $\sin(x)$ and $\cos(x)$. We know that when $\sin(x)$ is almost at it’s maximum, $\cos(x)$ is almost zero (at what value is $\sin(x)$ at its maximum?). Since any number divided by a very small number is very large, it should be obvious that your function will reach positive values in that interval.

J. W. Tanner
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Sina Babaei Zadeh
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When taking a test it is your job to eliminate any doubt about your abilities. If we cannot write clearly about simple matters, then we should not attempt to write about complicated matters.

Let $I = (-\frac{1}{2}, \frac{1}{2})$ and let $f : I \rightarrow \mathbb{R}$ be given by $$f(x) = \tan(\pi x).$$ Then $$f(x) \rightarrow -\infty, \quad x \rightarrow -\frac{1}{2}, \quad x \in I$$ and $$f(x) \rightarrow \infty, \quad x \rightarrow \frac{1}{2}, \quad x \in I.$$ Let $g : \mathbb{R} \rightarrow \mathbb{R}$ be given by $$g(x) = x + 6.$$ Then $$ g(x) \rightarrow \frac{11}{2}, \quad x \rightarrow -\frac{1}{2}, \quad x \in I$$ and $$ g(x) \rightarrow \frac{13}{2}, \quad x \rightarrow \frac{1}{2}, \quad x \in I.$$ It follows that $h : I \rightarrow \mathbb{R}$ given by $$ h(x) = f(x) - g(x)$$ satisfies $$ h(x) \rightarrow -\infty, \quad x \rightarrow -\frac{1}{2}, \quad x \in I$$ and $$ h(x) \rightarrow \infty, \quad x \rightarrow \frac{1}{2}, \quad x \in I.$$ In particular, there exists $x_1, x_2 \in I$, such that $h(x_1) < 0$ and $h(x_2) > 0$. The function $h$ is continuous, because $h$ is the difference of two continuous functions. By the intermediate value property $h$ has a zero in open interval between $x_1$ and $x_2$.
Carl Christian
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Hint : $$f(0.4)<0$$ $$f(0.47)>0$$ and in the interval $[0.4,0.47]$ , $f$ is continous, so this time, you can apply IVT.

Peter
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You can't use the intermediate value theorem between $0$ and $1$ as $tan(\pi{x})$ is undefined at $x=0.5$. However you can see that at $x=0$ it is negative. And at $x=0.5$ it is positive ($tan\frac{\pi}{2}$ is positive even though it is undefined). So using the intermediate value theorem between $0$ and $0.5$ we see that there is a root between it. So there is a root between $0$ and $1$ also.

J. W. Tanner
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Abhinav
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