## Thomas' Calculus 13th Edition

a. A root of f is the solution of $f(x)=0.$ This problem is to solve $\quad x^{3}-3x+1=0$, The solutions are numbers $c$ such that $f(c)=0.$ b. The problem is solved by equating the RHS's. $x^{3}=3x+1\ \Leftrightarrow\ x^{3}-3x+1=0$, which is the same problem as in a. We want $c$ such that $f(c)=0.$ c. The equation is equivalent to $\quad x^{3}-3x+1=0$, which is the same problem as in a. We want $c$ such that $f(c)=0.$ d. The problem is solved by equating the RHS's. $x^{3}-3x=1\ \Leftrightarrow\ x^{3}-3x+1=0$, which is the same problem as in a. We want $c$ such that $f(c)=0.$ e. This is the same problem as in a. We want $c$ such that $f(c)=0.$