Consider the function $p(q) =\dfrac{16}{q}$. The difference quotient of this function is equal to $4$ if $q$ decreases with $4$. Determine $q$.
$q=2$.
$q=-2$.
$q = 0 \mbox{ or } q=3$.
$q=2-2\sqrt{2} \mbox{ or } q=2+2\sqrt{2}$.
Consider the function $p(q) =\dfrac{16}{q}$. The difference quotient of this function is equal to $4$ if $q$ decreases with $4$. Determine $q$.
Antwoord 1 correct
Correct
Antwoord 2 optie
$q=-2$.
Antwoord 2 correct
Fout
Antwoord 3 optie
$q = 0 \mbox{ or } q=3$.
Antwoord 3 correct
Fout
Antwoord 4 optie
$q=2-2\sqrt{2} \mbox{ or } q=2+2\sqrt{2}$.
Antwoord 4 correct
Fout
Antwoord 1 optie
$q=2$.
Antwoord 1 feedback
Correct: The difference quotient is
$$
\begin{align*}
\dfrac{\Delta p}{\Delta q} &= \dfrac{p(q-4)-p(q)}{-4} = \dfrac{\tfrac{16}{q-4}-\tfrac{16}{q}}{-4} = \dfrac{1}{-4}\cdot\left(\dfrac{16}{q-4}-\dfrac{16}{q}\right) = \dfrac{\tfrac{16}{-4}}{q-4}-\dfrac{\tfrac{16}{-4}}{q} \\
&= \dfrac{-4}{q-4}-\dfrac{-4}{q} = \dfrac{-4q}{(q-4)q} + \dfrac{4(q-4)}{q(q-4)} = \dfrac{4(q-4)-4q}{q(q-4)} = \dfrac{4q-16 - 4q}{q^2-4q} = \dfrac{-16}{q^2-4q}
\end{align*}
$$
Solving $\dfrac{\Delta p}{\Delta q} = \dfrac{-16}{q^2-4q}=4$ gives
$$
\begin{align*}
\dfrac{-16}{q^2-4q}&=4\\
-16 &= 4(q^2-4q) = 4q^2 - 16 q\\
-4q^2 + 16q - 16&= 0\\
q^2 -4q + 4 &= 0\\
(q-2)^2 &= 0\\
q-2 &= 0\\
q &= 2.
\end{align*}
$$

Go on.
Antwoord 2 feedback
Wrong: Note the sign of $\Delta q$; there is a decrease of $4$, hence $\Delta q = -4$.

See also Difference quotient and Example 1.
Antwoord 3 feedback
Wrong: It is given that the change in $q$ equals $-4$, not that the start value of $q$ equals $-4$.

See also Example 2.
Antwoord 4 feedback
Wrong: Note the minus-signs while working out the differnce quotient.

See also Example 2.