Chapter 4: Problem 20
Given \(f(x)=x^{3}-1, \quad x>1\) \(=x-1, \quad x \leq 1\), find f(1)
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
/*! 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 20
Given \(f(x)=x^{3}-1, \quad x>1\) \(=x-1, \quad x \leq 1\), find f(1)
These are the key concepts you need to understand to accurately answer the question.
All the tools & learning materials you need for study success - in one app.
Get started for free
Given \(\begin{aligned} f(x) &=a x(x-1)+b, \quad x<1 \\ &=x-1, \quad 1 \leq x \leq 3 \\ &=p x^{2}+q x+2, \quad x>3 \end{aligned}\) Find the constants \(a, b, p\) and \(q\) so that \(f(x)\) is differentiable at \(x=1 \& x=3\). \(\left.p=\frac{1}{3}, q=-1\right\\}\)
$$ \left\\{\begin{array}{l} x=a \cos ^{3} t \\ y=b \sin ^{3} t \end{array}\right. $$
$$ y=-\frac{x}{1+8 x^{3}}+\frac{1}{12} \ln \frac{(1+2 x)^{2}}{1-2 x+4 x^{2}}+\frac{\sqrt{3}}{6} \tan ^{-1} \frac{4 x-1}{\sqrt{3}} $$
$$ \text { If } y=e^{2 x+3}\left(x^{2}-x+\frac{1}{2}\right), \text { find } \frac{d y}{d x},\left(\frac{d y}{d x}\right)_{x=0} $$
What do you think about this solution?
We value your feedback to improve our textbook solutions.