Chapter 5: Integrals - Section 5.4 - The Fundamental Theorem of Calculus - Exercises 5.4 - Page 288: 70

$L(x)=-2x+1$

The function $f(x)$ is the derivative of $f(x)=3+\int_1^{x^2} \sec (t-1) dt$ is equal to $g'(x)=2x\sec(x^2-1)$ by the fundamental Theorem of calculus. Here,we have $g(1)=3$ and $ g'(1)=-2$ The linearization of $f(x)$ at $x=-1$ is gievn as: $L(x)=g(-1)+g'(-1)(x+1)$ $\implies L(x) =3-2(x+1)$ Thus, $L(x)=-2x+1$