Certainly! Let's solve for the roots of the linear function [tex]\( f(x) = a x + b \)[/tex]. Here's a detailed, step-by-step approach:
1. Understand the problem: We are given a linear function [tex]\( f(x) = a x + b \)[/tex] and asked to find the value of [tex]\( x \)[/tex] where this function equals zero. This is equivalent to finding the solution of the equation [tex]\( a x + b = 0 \)[/tex], with the condition that [tex]\( a \neq 0 \)[/tex].
2. Set up the equation: To find the root of the function, we set the function equal to zero:
[tex]\[
f(x) = 0 \implies a x + b = 0
\][/tex]
3. Isolate [tex]\( x \)[/tex]: We want to solve the equation for [tex]\( x \)[/tex]. Start by isolating the term involving [tex]\( x \)[/tex] on one side of the equation:
[tex]\[
a x + b = 0
\][/tex]
Subtract [tex]\( b \)[/tex] from both sides:
[tex]\[
a x = -b
\][/tex]
4. Solve for [tex]\( x \)[/tex]: Since [tex]\( a \neq 0 \)[/tex], we can safely divide both sides of the equation by [tex]\( a \)[/tex] to solve for [tex]\( x \)[/tex]:
[tex]\[
x = -\frac{b}{a}
\][/tex]
Therefore, the root of the function [tex]\( f(x) = a x + b \)[/tex] is given by:
[tex]\[
x = -\frac{b}{a}
\][/tex]
This result shows that for any non-zero value of [tex]\( a \)[/tex], the function [tex]\( f(x) = a x + b \)[/tex] has a unique root at [tex]\( x = -\frac{b}{a} \)[/tex].
