Exercise 20

Consider the function whose correspondence rule is \( f(x) = x^3 - x \). If the line with equation \( y = mx + n \) cuts the graph of this function at points \( ( 0, \, f(0) ) \) and \( ( 1, \, f(1) ) \), then we can say of \( m \) and \( n \) that:

  1. both are positive

  2. both are negative

  3. both are null

  4. \(m\) is null and \(n\) is not null

  5. \(m\) is not null and \(n\) is null

Try to solve it before checking the answer.

  1. both are null

We have that:

\[ f(0) = 0^3 – 0 = 0 \]

and

\[ f(1) = 1^3 – 1 = 0 \]

Luego, los puntos de intersección son \(( 0, \, 0 )\)   y   \( (1, \, 0) \). Las coordenadas de estos puntos deben satisfacer la ecuación de la recta. Luego:

Then, the points of intersection are (0,0) and (1,0). The coordinates of these points must satisfy the equation of the line. Now:

\[ \begin{aligned} &0 = m(0) + n \quad \text{ and } \quad 0 = m(1) + n \\[.5em] &\hspace{3em} \Rightarrow 0 = n \quad \text{ and } \quad 0 = m + n \\[.5em] &\hspace{3em}\Rightarrow \boldsymbol{ n = 0 \quad \text{ and } \quad m = 0 } \end{aligned} \]
\[ \begin{aligned} 0 = m(0) + n \quad \text{ and } \quad 0 = m(1) + n &\Rightarrow 0 = n \quad \text{ and } \quad 0 = m + n \\[.5em] &\Rightarrow \boldsymbol{ n = 0 \quad \text{ and } \quad m = 0 } \end{aligned} \]

Hence, both are null.

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