Chapter 3: Problem 59
Show that the equation \(x^{5}-3 x-1=0\) has at least one root lying between 1 and 2 .
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Chapter 3: Problem 59
Show that the equation \(x^{5}-3 x-1=0\) has at least one root lying between 1 and 2 .
These are the key concepts you need to understand to accurately answer the question.
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Given \(\begin{aligned} f(x) &=x^{2}+1, &-2 \leq x<0 \\ &=-x^{2}-1, & 0 & \leq x \leq 2 . \end{aligned}\) Is there a point in the interval \([-2,2]\) at which \(f(x)=0 ?\)
Choose \(A\) and \(B\) so as to make the function \(f(x)\) continuous at \(x=\pm
\frac{\pi}{2}\)
$$
\begin{aligned}
f(x) &=-2 \sin x, \quad x \leq-\frac{\pi}{2} \\
&=A \sin x+B, \quad-\frac{\pi}{2}
If \(f(x y)=f(x)+f(y) \forall x, y \neq 0\) and \(f(x)\) is continuous at \(x=1\), then check the continuity of \(f(x)\).
If \(f(x+2 y)=f(x)[f(y)]^{2} \forall x, y\) and \(f(x)\) is continuous at \(x=0\), then check the continuity of \(f(x)\).
If \(f(x+y)=f(x)+f(y) \forall x, y\) and \(f(x)\) is continuous at \(x=0\), then show that \(f(x)\) is continuous \(\forall x\).
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