Thomas Calculus 13th Edition Exercise 1.1 Solution 'link'

( g(-x) = (-x)^2 + (-x) = x^2 - x ). This is neither ( g(x) ) nor ( -g(x) ). Answer: Neither.

Algebraic check ( f(-x) = f(x) ) or ( f(-x) = -f(x) ) is performed properly for functions like ( f(x) = x^3 - x ).

( (f+g)(x) = (x+1)+(x-1) = 2x ). Domain: all reals. thomas calculus 13th edition exercise 1.1 solution

Starting from ( y = x^3 ), reflect over x-axis then shift up 3.

If you need deeper explanations or additional problems, several platforms provide comprehensive walkthroughs: ( g(-x) = (-x)^2 + (-x) = x^2 - x )

Before diving into solutions, let’s review the core objectives of this exercise:

Prove if f and g are even, then f+g is even. Algebraic check ( f(-x) = f(x) ) or

Vertex at ( x = -b/(2a) = -4/(2*(-1)) = 2 ). ( f(2) = -4 + 8 - 1 = 3 ). Parabola opens downward → range = ( (-\infty, 3] ).