AP Calculus BC 2.2 Defining the Derivative of a Function and Using Derivative Notation - FRQs - Exam Style Questions
No-Calc Question
Most-appropriate topic codes (CED):
• TOPIC 7.1: Modeling Situations with Differential Equations — part (b)
• TOPIC 5.4: Using the First Derivative Test to Determine Relative (Local) Extrema — part (c)
• TOPIC 2.2: Defining the Derivative of a Function and Using Derivative Notation — part (d)
▶️ Answer/Explanation
(a)
Sample slope field at the six indicated points:
(b)
\(\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{d}{dx}(2x-y)=2-\frac{dy}{dx}=2-(2x-y)=2-2x+y.\)
In Quadrant II, \(x<0\) and \(y>0\). Then \(2-2x+y>0\), so \(\displaystyle \frac{d^{2}y}{dx^{2}}>0\).
Therefore, all solution curves are \(\boxed{\text{concave up in Quadrant II}}\).
(c)
\(f'(x)=2x-f(x)\). At \((x,y)=(2,3)\): \(f'(2)=2(2)-3=1\ne0\).
Since the derivative is not zero, \(f\) has \(\boxed{\text{neither a relative minimum nor a relative maximum at }x=2}\).
(d)
Suppose \(y=mx+b\). Then \(\dfrac{dy}{dx}=m\). Substitute into \(y’=2x-y\):
\(m=2x-(mx+b)\Rightarrow (2-m)x-(m+b)=0\ \text{for all }x.\)
Hence \(2-m=0\Rightarrow m=2\) and \(-(m+b)=0\Rightarrow b=-m=-2\).
\(\boxed{m=2,\; b=-2}\).