AP Calculus BC 3.6 Calculating Higher- Order Derivatives - FRQs - Exam Style Questions

(a) Compute \(y”\) in terms of \(x\) and \(y\)
From \(y’ = x^2 – \tfrac{1}{2}y\), differentiate: \(y” = 2x – \tfrac{1}{2}y’\).
Substitute \(y’\): \(y” = 2x – \tfrac{1}{2}\!\left(x^2 – \tfrac{1}{2}y\right) = 2x – \tfrac{1}{2}x^2 + \tfrac{1}{4}y.\)

(b) Classify the extremum at \((-2,8)\)
\(y'(-2) = (-2)^2 – \tfrac{1}{2}(8) = 4 – 4 = 0.\)
Using the result from (a): \(y”(-2,8) = 2(-2) – \tfrac{1}{2}(-2)^2 + \tfrac{1}{4}(8) = -4 – 2 + 2 = -4 < 0.\)
Since \(y'(-2)=0\) and \(y”(-2)<0\), \(f\) has a relative maximum at \((-2,8)\).

(c) Limit using L’Hospital’s Rule
Given \(g(-1)=2\), we have \(\displaystyle \lim_{x\to -1}\bigl(g(x)-2\bigr)=0\) and \(\displaystyle \lim_{x\to -1}3(x+1)^2=0\) \(\Rightarrow\) indeterminate \(0/0\).
First application: \[ \lim_{x\to -1}\frac{g(x)-2}{3(x+1)^2} = \lim_{x\to -1}\frac{g'(x)}{6(x+1)}. \] Since \(g'(x)=x^2-\tfrac{1}{2}g(x)\), \(g'(-1)=1-\tfrac{1}{2}\cdot 2=0\) and denominator \(\to 0\), so apply L’Hospital again:
\[ \lim_{x\to -1}\frac{g'(x)}{6(x+1)}=\lim_{x\to -1}\frac{g”(x)}{6}=\frac{g”(-1)}{6}. \] Using (a) with \(g(-1)=2\): \(g”(-1)=2(-1)-\tfrac{1}{2}(-1)^2+\tfrac{1}{4}\cdot 2=-2-\tfrac{1}{2}+ \tfrac{1}{2}=-2.\)
Therefore, \[ \lim_{x\to -1}\frac{g(x)-2}{3(x+1)^2}=\frac{-2}{6}=\boxed{-\tfrac{1}{3}}. \]

(d) Euler’s method, two equal steps from \(x=0\) to \(x=1\)
Step size \(h=\tfrac{1}{2}\). Given \(h(0)=2\).
Slope at \(x=0\): \(h'(0)=0^2-\tfrac{1}{2}\cdot 2=-1\).
First step to \(x=\tfrac{1}{2}\): \(h\!\left(\tfrac{1}{2}\right)\approx h(0)+h\,'(0)\,h=2+(-1)\cdot \tfrac{1}{2}=\tfrac{3}{2}.\)
Slope at \(x=\tfrac{1}{2}\): \(h’\!\left(\tfrac{1}{2}\right)=\left(\tfrac{1}{2}\right)^2-\tfrac{1}{2}\cdot \tfrac{3}{2}=\tfrac{1}{4}-\tfrac{3}{4}=-\tfrac{1}{2}.\)
Second step to \(x=1\): \(h(1)\approx h\!\left(\tfrac{1}{2}\right)+h’\!\left(\tfrac{1}{2}\right)\,h=\tfrac{3}{2}+ \left(-\tfrac{1}{2}\right)\!\cdot \tfrac{1}{2}=\tfrac{5}{4}.\)
So \(h(1)\approx \boxed{\tfrac{5}{4}}.\)