Applications of Derivatives
Application of Derivatives
nta_pyq_2025_apr
Grade 12
Question:
The sum of all local minimum values of the function
$$f(x) = \begin{cases} 1-2x, & x < -1 \\ \dfrac{1}{3}(7+2|x|), & -1 \leq x \leq 2 \\ \dfrac{11}{18}(x-4)(x-5), & x > 2 \end{cases}$$
is
$\dfrac{157}{72}$
$\dfrac{131}{72}$
$\dfrac{171}{72}$
$\dfrac{167}{72}$
Step-by-Step Solution
Key Concept: Analyse each piece separately: for $-1 \leq x \leq 2$, note $f(x) = \tfrac{1}{3}(7+2|x|)$ has a local min at $x=0$ (value $\tfrac{7}{3}$) and check the junction $x=2$; for $x>2$, find the vertex of the quadratic $\tfrac{11}{18}(x-4)(x-5)$.
Piece 1 ($x<-1$): $f(x) = 1-2x$ is decreasing, no local min.
Middle piece ($-1 \leq x \leq 2$): $f(x) = \tfrac{1}{3}(7+2|x|)$. At $x=0$, this has a local minimum $f(0) = \tfrac{7}{3}$.
Piece 3 ($x>2$): $f(x) = \tfrac{11}{18}(x-4)(x-5)$. Setting $f'(x) = \tfrac{11}{18}(2x-9) = 0 \Rightarrow x = \tfrac{9}{2}$. Local minimum at $x = \tfrac{9}{2}$:
$$f\!\left(\frac{9}{2}\right) = \frac{11}{18}\left(-\frac{1}{2}\right)\!\left(-\frac{1}{2}\right) = -\frac{11}{72}.$$
Sum of local minima $= \dfrac{7}{3} - \dfrac{11}{72} = \dfrac{168-11}{72} = \dfrac{157}{72}$.
Correct Answer: 1