This question assesses the connection between differentiability and continuity in AP Calculus AB, exploring piecewise functions and their derivatives at join points. For the given piecewise function, it is continuous at x=0 since both pieces approach 0. However, the left-hand derivative is 0 (from 2x at x=0-), while the right-hand derivative is -1 (from -x). This difference means the derivative does not exist at x=0. A tempting distractor might say yes because of continuity, but equal one-sided derivatives are also required. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.

This question explores the implication of differentiability on continuity. If f is differentiable at x=7, then by theorem, it must be continuous there, as differentiability requires the limit of the difference quotient to exist, which implies continuity. The existence of f'(7) ensures the function approaches the same value from both sides at x=7. No exceptions exist where a function is differentiable but discontinuous. Choice A is a tempting distractor, claiming differentiability does not imply continuity, but it fails because the theorem states the opposite. To check differentiability at a point, verify continuity there and ensure the left- and right-hand derivatives exist and are equal.

This question probes differentiability versus continuity for piecewise linear functions. For this f(x), at x=2, the function is continuous since both pieces meet at 3, but the left slope is 1 and the right is 3. These differing one-sided slopes mean f'(2) does not exist. Continuity is present, but differentiability requires matching derivatives from both sides. Choice C is a tempting distractor, stating continuity ensures the derivative exists, but it fails because unequal slopes create a corner. To check differentiability at a point, verify continuity there and ensure the left- and right-hand derivatives exist and are equal.

This question assesses the connection between differentiability and continuity in AP Calculus AB, examining functions with absolute values and roots. For f(x) = $|x-1|^{1/2}$, it is continuous at x=1 with value 0. However, the left-hand derivative approaches -infinity, and the right-hand approaches +infinity, creating a vertical tangent. This infinite mismatch means f'(1) does not exist. A tempting distractor might say yes because of continuity, but finite matching slopes are required. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.

This question tests differentiability versus continuity in cases of vertical tangents. A vertical tangent at x=0 means the slope is infinite, so the limit of the difference quotient does not exist finitely. Therefore, f'(0) does not exist, even if f is continuous there. Continuity is required but not sufficient when the derivative is infinite. Choice A is a tempting distractor, claiming continuity implies differentiability, but it fails because infinite slopes prevent it. To check differentiability at a point, verify continuity there and ensure the left- and right-hand derivatives exist and are equal.

This question assesses the connection between differentiability and continuity in AP Calculus AB, analyzing piecewise linear functions. For this f(x), it is continuous at x=0 with value 2 matching both limits. However, the left-hand derivative is 1, while the right-hand is -1, creating a corner. This difference means f'(0) does not exist. A tempting distractor might say yes because of continuity, but slopes must match. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.

This question evaluates the connection between differentiability and continuity for piecewise functions. For this f(x), at x=0, the left-hand limit is 0 but f(0)=1, creating a discontinuity. Since differentiability implies continuity, the lack of continuity means f'(0) does not exist. The pieces may be differentiable individually, but the jump at x=0 breaks continuity. Choice C is a tempting distractor, claiming continuity at x=0 ensures the derivative exists, but it fails because the function is actually discontinuous there. To check differentiability at a point, verify continuity there and ensure the left- and right-hand derivatives exist and are equal.

This question assesses the connection between differentiability and continuity in AP Calculus AB, considering redefined points in otherwise continuous functions. For this f(x), lim x→0 sin x = 0, but f(0)=1, creating a discontinuity. Differentiability requires continuity, so f'(0) cannot exist here. The redefinition breaks the necessary limit agreement. A tempting distractor might say yes because redefining one point doesn't affect differentiability, but it does if it causes discontinuity. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.

This question assesses the connection between differentiability and continuity in AP Calculus AB, focusing on how certain function behaviors affect the derivative. For f(x) = √x, the function is continuous at x=0 since the limit and function value are both 0. However, the right-hand derivative at x=0 is lim h→0+ (√h - 0)/h = lim 1/√h, which approaches infinity. Since the slope becomes infinitely steep, creating a vertical tangent, the derivative does not exist at x=0. A tempting distractor might say yes because of continuity, but continuity alone does not ensure differentiability when the slope is undefined. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.

This question examines the link between differentiability and continuity, specifically regarding vertical tangents. A vertical tangent at x=5 indicates that the slope approaches infinity, meaning the limit of the difference quotient does not exist as a finite number. For p'(5) to exist, this limit must be finite and equal from both sides, but an infinite slope prevents that. The function can be continuous at x=5 with a vertical tangent, yet differentiability requires a finite derivative. Choice C is a tempting distractor, stating continuity ensures the derivative exists, but it fails because vertical tangents make the derivative undefined despite continuity. To check differentiability at a point, verify continuity there and ensure the left- and right-hand derivatives exist and are equal.