IB DP Math MAI SL : IB Style Mock Exams – Set 4 Paper 1
Question
(i) the local maximum point.
(ii) the local minimum point.
(i) the gradient of this tangent.
(ii) the equation of this tangent, giving your answer in the form \( y = mx + c \).
▶️ Answer/Explanation
(a)(i)
The local maximum occurs where \( f'(x) = 0 \) and the second derivative is negative. From the graph or using calculus: \( f'(x) = 3x^2 + 4x – 4 \). Solving \( f'(x) = 0 \) gives critical points. The local maximum is at \( x = -2 \).
\( \boxed{-2} \).
(a)(ii)
The local minimum is at \( x = \frac{2}{3} \) (or approximately 0.667).
\( \boxed{\frac{2}{3}} \) (or 0.667).
(b)
The function is decreasing where \( f'(x) < 0 \), which is between the two turning points.
\( \boxed{-2 < x < \frac{2}{3}} \).
(c)(i)
The tangent at \( (1, -3) \) is parallel to \( y = 3x + 5 \), so its gradient is the same as the slope of that line, which is 3.
\( \boxed{3} \).
(c)(ii)
Using point-gradient form: \( y – y_1 = m(x – x_1) \). With \( m = 3 \) and point \( (1, -3) \):
\( y + 3 = 3(x – 1) \)
\( y = 3x – 6 \).
\( \boxed{y = 3x – 6} \) or equivalent.