Transforming Equations
Consider the equation g(x) = h(x) and the single-valued function f (x). A particular value of x is a solution of the equation if substituting that value into the equation results in an identity. In determining the solutions of an equation, we often apply functions to each side of the equation in order to simplify its form. We apply the function to obtain a second equation, f (g(x)) = f (h(x)). If x = ξ is a solution of the former equation, (let ψ = g(ξ) = h(ξ)), then it is necessarily a solution of latter. This is because f (g(ξ)) = f (h(ξ)) reduces to the identity f (ψ) = f (ψ). If f (x) is bijective, then the converse is true: any solution of the latter equation is a solution of the former equation. Suppose that x = ξ is a solution of the latter, f (g(ξ)) = f (h(ξ)). That f (x) is a one-to-one mapping implies that g(ξ) = h(ξ). Thus x = ξ is a solution of the former equation.
Sumber
Consider the equation g(x) = h(x) and the single-valued function f (x). A particular value of x is a solution of the equation if substituting that value into the equation results in an identity. In determining the solutions of an equation, we often apply functions to each side of the equation in order to simplify its form. We apply the function to obtain a second equation, f (g(x)) = f (h(x)). If x = ξ is a solution of the former equation, (let ψ = g(ξ) = h(ξ)), then it is necessarily a solution of latter. This is because f (g(ξ)) = f (h(ξ)) reduces to the identity f (ψ) = f (ψ). If f (x) is bijective, then the converse is true: any solution of the latter equation is a solution of the former equation. Suppose that x = ξ is a solution of the latter, f (g(ξ)) = f (h(ξ)). That f (x) is a one-to-one mapping implies that g(ξ) = h(ξ). Thus x = ξ is a solution of the former equation.
It is always safe to apply a
one-to-one, (bijective), function to an equation, (provided it is defined for
that domain). For example, we can apply f (x) = x3
or f (x) = ex, considered as mappings on R, to
the equation x = 1. The equations x3 = 1 and ex
= e each have the unique solution x = 1 for x ∈R.
In general, we must take care in
applying functions to equations. If we apply a many-to-one function, we may
introduce spurious solutions. Applying f (x) = x2
to the equation x = π/2 results in 
x = {± π/2}. Applying f
(x) = sin x results in , which has the two solutions, which has an infinite number of solutions, 
x = {± π/2}. Applying f
(x) = sin x results in , which has the two solutions, which has an infinite number of solutions,
We do not generally apply a
one-to-many, (multi-valued), function to both sides of an equation as this
rarely is useful. Rather, we typically use the definition of the inverse
function. Consider the equation
sin2 x
= 1.
Applying the function f (x) = x1/2
to the equation would not get us anywhere.
(sin2 x)1/2
= 11/2
Since (sin2 x)1/2
≠
sin x, we cannot simplify the left side of the equation. Instead we
could use the definition off (x) = x1/2 as the inverse
of the x2 function to obtain
sin x = 11/2
= (±1).
Now note that we should not just
apply arcsin to both sides of the equation as arcsin(sin x) ≠ x.
Instead we use the definition of arcsin as the inverse of sin.
x = arcsin (±1)
x = arcsin (1) has the solutions x = π/2+2nπ and x
= arcsin (−1) has the solutions x = −π/2 + 2nπ. We enumerate the
solutions.
x = {nπ
2 + nπ | n ∈Z}.
Sumber
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