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

Question

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?

Answer 1

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.

Answer 2

$\tan(x)$ 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.

Answer 3

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.

Answer 4

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.

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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 $x_2$.