Let the function ggg be defined by g(x)={x2−2x+1if x<1ex−1−1if x≥1\displaystyle g(x) = \begin{cases} \displaystyle x^2 - 2x + 1 & \text{if } x < 1 \\ \displaystyle e^{x-1} - 1 & \text{if } x \geq 1 \end{cases}g(x)={x2−2x+1ex−1−1if x<1if x≥1. Which of the following statements about ggg are true?
I. limx→1−g(x)=limx→1+g(x)\displaystyle \lim_{x \to 1^-} g(x) = \lim_{x \to 1^+} g(x)x→1−limg(x)=x→1+limg(x)
II. limx→1−g′(x)=limx→1+g′(x)\displaystyle \lim_{x \to 1^-} g'(x) = \lim_{x \to 1^+} g'(x)x→1−limg′(x)=x→1+limg′(x)
III. ggg is differentiable at x=1x = 1x=1.
I only
None
I and II only
III only
Not enough data yet to show leaderboard.
AI Tutor 3:10 AM
How can I help?