AP Calculus BC 2.7: Derivative-Definition Limits

Because f is differentiable at x=2, f is continuous at x=2, so $\lim _{x \rightarrow 2} f(x)=f(2)=1$.
$$\begin{aligned} \lim _{x \rightarrow 2}(6 f(x)-3 x)=6 \cdot 1-3 \cdot 2=0 \lim _{x \rightarrow 2}\left(x^{2}-5 x+6\right)=0 \end{aligned}$$

Limits of numerator and denominator

1 point

Uses L'Hospital's
Rule

1 point

Because $\lim _{x \rightarrow 2} \frac{6 f(x)-3 x}{x^{2}-5 x+6}$ is of indeterminate form $\frac{0}{0}$,

L'Hospital's Rule can be applied.

Using L'Hospital's Rule,
$$\lim _{x \rightarrow 2} \frac{6 f(x)-3 x}{x^{2}-5 x+6}=\lim _{x \rightarrow 2} \frac{6 f^{\prime}(x)-3}{2 x-5}=\frac{6 \cdot 0-3}{2 \cdot 2-5}=3.$$

Answer

1 point

Scoring notes:

- The first point is earned by the presentation of two separate limits for the numerator and

denominator.

- A response that presents a limit explicitly equal to $\frac{0}{0}$ does not earn the first point.

- The second point is earned by applying L'Hospital's Rule, that is, by presenting at least one correct

derivative in the limit of a ratio of derivatives.

- The third point is earned for the correct answer with supporting work.

Total for part (c)

3 points

(d) Find the absolute minimum value of f on the closed interval [-2, 8]. Justify your answer.