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  2. For business economics
  3. Chapter 2: Differentiation of functions of one variable
  4. Derivative
  5. Difference quotient
  6. Exercise 4

Exercise 4

Consider the function $y(x) = (x+3)^2$. The difference quotient of this function is equal to $-1$ at $x=0$. Determine $\Delta x$.
$\Delta x = -7$.
This cannot be solved, as it is not allowed to divide by zero.
$\Delta x = -1$.
$\Delta x = -5$.
Consider the function $y(x) = (x+3)^2$. The difference quotient of this function is equal to $-1$ at $x=0$. Determine $\Delta x$.
Antwoord 1 correct
Correct
Antwoord 2 optie
This cannot be solved, as it is not allowed to divide by zero.
Antwoord 2 correct
Fout
Antwoord 3 optie
$\Delta x = -1$.
Antwoord 3 correct
Fout
Antwoord 4 optie
$\Delta x = -5$.
Antwoord 4 correct
Fout
Antwoord 1 optie
$\Delta x = -7$.
Antwoord 1 feedback
Correct: The difference quotient is $$\dfrac{\Delta y}{\Delta x} = \dfrac{y(0+\Delta x)-y(0)}{\Delta x} = \dfrac{(\Delta x + 3)^2 - 9}{\Delta x} = \dfrac{(\Delta x)^2+6\Delta x + 9 - 9}{\Delta x} = \dfrac{(\Delta x)^2+6\Delta x}{\Delta x} = \Delta x + 6 = -1.$$
Solving this gives $\Delta x = -7$.

Go on.
Antwoord 2 feedback
Wrong: The start value of $x$ is given, not the change in $x$.

See also Example 2.
Antwoord 3 feedback
Wrong: Pay attention when working out brackets. $(\Delta x + 3)^2 \neq (\Delta x)^2+9$.

See also Example 2.
Antwoord 4 feedback
Wrong: Pay attention to the order of $y(x + \Delta x)$ and $y(x)$.

See also Example 2.