Chapter 4: Problem 18
Show that the cubic \(2 x^{3}+3 x^{2}+6 x+10\) has exactly one real root.
/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none}
Learning Materials
Features
Discover
Chapter 4: Problem 18
Show that the cubic \(2 x^{3}+3 x^{2}+6 x+10\) has exactly one real root.
All the tools & learning materials you need for study success - in one app.
Get started for free
Use the MVT to show that \(10.049<\sqrt{101}<10.05\) and \(10.24<\sqrt{105}<\) 10.25. Also, find better estimates using Taylor's Theorem with \(n=1\), that is, using the Extended MVT.
Let \(I\) be an interval containing more than one point and \(f: I \rightarrow \mathbb{R}\) be any function. (i) Assume that \(f\) is differentiable. If \(f^{\prime}\) is nonnegative on \(I\) and \(f^{\prime}\) vanishes at only a finite number of points on any bounded subinterval of \(I\), then show that \(f\) is strictly increasing on \(I\). (ii) Assume that \(f\) is twice differentiable. If \(f^{\prime \prime}\) is nonnegative on \(I\) and \(f^{\prime \prime}\) vanishes at only a finite number of points on any bounded subinterval of \(I\), then show that \(f\) is strictly convex on \(I\). (iii) Consider \(f: \mathbb{R} \rightarrow \mathbb{R}\) given by \(f(x)=(x-2 n)^{3}+2 n\), where \(n \in \mathbb{Z}\) is such that \(x \in[2 n-1,2 n+1) .\) Show that \(f\) is differentiable on \(\mathbb{R}\) and \(f^{\prime \prime}\) exists on \((2 n-1,2 n+1)\), but \(f_{+}^{\prime \prime}(2 n+1)=6\), whereas \(f_{-}^{\prime \prime}(2 n+1)=-6\) for each \(n \in \mathbb{N} .\) Also show that \(f\) is strictly increasing on \(\mathbb{R}\) although \(f^{\prime}(2 n)=0\) for each \(n \in \mathbb{N}\). (Compare (i) above and Exercise 12 in the list of Revision Exercises at the end of Chapter 7.) (iv) Consider \(g: \mathbb{R} \rightarrow \mathbb{R}\) given by \(g(x)=(x-2 n)^{4}+8 n x\), where \(n \in \mathbb{Z}\) is such that \(x \in[2 n-1,2 n+1) .\) Show that \(g\) is twice differentiable on \(\mathbb{R}\) and \(g^{\prime \prime \prime}\) exists on \((2 n-1,2 n+1)\), but \(g_{+}^{\prime \prime \prime}(2 n+1)=24\), whereas \(g_{-}^{\prime \prime \prime}(2 n+1)=-24\) for each \(n \in \mathbb{N} .\) Also show that \(g\) is strictly convex on \(\mathbb{R}\) although \(g^{\prime \prime}(2 n)=0\) for each \(n \in \mathbb{N}\). (Compare (ii) above and Exercise 13 in the list of Revision Exercises at the end of Chapter \(7 .\) )
Let a function \(f:[a, b] \rightarrow \mathbb{R}\) be continuous and its second
derivative \(f^{\prime \prime}\) exist everywhere on the open interval \((a, b)
.\) Suppose the line segment joining \((a, f(a))\) and \((b, f(b))\) intersects the
graph of \(f\) at a third point (c, \(f(c))\), where \(a
Let \(a>0\) and \(f:[-a, a] \rightarrow \mathbb{R}\) be continuous. Suppose \(f^{\prime}(x)\) exists and \(f^{\prime}(x) \leq 1\) for all \(x \in(-a, a)\). If \(f(a)=a\) and \(f(-a)=-a\), then show that \(f(x)=x\) for every \(x \in(-a, a)\).
Find the \(n\) th Taylor polynomial of \(f\) around \(a\), that is, $$ P_{n}(x)=f(a)+f^{\prime}(a)(x-a)+\cdots+\frac{f^{(n)}(a)}{n !}(x-a)^{n} \quad \text { for } x \in \mathbb{R} $$ when \(a=0\) and \(f(x)\) equals: (i) \(\frac{1}{1-x}\), (ii) \(\frac{1}{1+x}\), (iii) \(\frac{x}{1+x^{2}}\).
What do you think about this solution?
We value your feedback to improve our textbook solutions.