Consider the function $y(x) = x^3 + 5$. The difference quotient of this function is $7$ at an increase of $x$ by $\Delta x=1$. Determine $x$.
Antwoord 1 correct
Correct
Antwoord 2 optie
$x=1$.
Antwoord 2 correct
Fout
Antwoord 3 optie
There is no solution, because solving results in the equation $1=7$.
Antwoord 3 correct
Fout
Antwoord 4 optie
$x=-4 \mbox{ or } x=1$.
Antwoord 4 correct
Fout
Antwoord 1 optie
$x=-2$ or $x=1$.
Antwoord 1 feedback
Correct: The difference quotient is
$$
\begin{align*}
\dfrac{\Delta y}{\Delta x} &= \dfrac{y(x+1)-y(x)}{1}\\
&= (x+1)^3 +5 - (x^3 + 5)\\
&= x^3 + 3x^2 + 3x + 1+ 5 - x^3 - 5\\
&= 3x^2 + 3x + 1\\
&= 7.
\end{align*}
$$
Hence,
$$
\begin{align*}
3x^2 + 3x - 6 &= 0\\
x^2 + x - 2 &= 0\\
(x+2)(x-1) &= 0\\
x + 2 =0 &\mbox{ or } x-1 = 0\\
x=-2 &\mbox{ or } x = 1.
\end{align*}
$$
Hence, there are two possible starting values for $x$; $x=-2$ or $x=1$.
$$
\begin{align*}
\dfrac{\Delta y}{\Delta x} &= \dfrac{y(x+1)-y(x)}{1}\\
&= (x+1)^3 +5 - (x^3 + 5)\\
&= x^3 + 3x^2 + 3x + 1+ 5 - x^3 - 5\\
&= 3x^2 + 3x + 1\\
&= 7.
\end{align*}
$$
Hence,
$$
\begin{align*}
3x^2 + 3x - 6 &= 0\\
x^2 + x - 2 &= 0\\
(x+2)(x-1) &= 0\\
x + 2 =0 &\mbox{ or } x-1 = 0\\
x=-2 &\mbox{ or } x = 1.
\end{align*}
$$
Hence, there are two possible starting values for $x$; $x=-2$ or $x=1$.
Antwoord 2 feedback
Wrong: $x=1$ is indeed a correct starting value, but there is another one.
See also Extra explanation quadratic functions: zeros.
See also Extra explanation quadratic functions: zeros.
Antwoord 3 feedback
Wrong: Pay attention to the working out of brackets. $(\Delta x + 1)^3 \neq (\Delta x)^3+1$.
See also Example 2.
See also Example 2.
Antwoord 4 feedback