## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 3: Derivatives - Section 3.1 - Tangents and the Derivative at a Point - Exercises 3.1 - Page 109: 33

#### Answer

$(x_0,mx_0+b)$

#### Work Step by Step

As we know that $y=mx+b$ Now, at any point $(x_0,mx_0+b)$, we have: $y'=\lim\limits_{h\to0}\dfrac{m(x_0+h)+b-(mx_0+b)}{h}=\lim\limits_{h\to0}\dfrac{mx_0+mh+b-mx_0-b}{h}=\lim\limits_{h\to0}\dfrac{mh}{h}=m$ Thus, the tangent line at any point $x=x_0$ can be written as:$y=mx+n$ Next, consider point $(x_0,mx_0+b)$: $mx_0+n=mx_0+b \implies n=b$ So, the tangent line of the given line is $y=mx+b$ will be remained as the same line. Hence, this implies that the line $y=mx+b$ has its own tangent line at the point $(x_0,mx_0+b)$.

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