This question directly addresses what unequal one-sided derivatives mean for differentiability and continuity. If left and right derivatives at x = 1 exist but are unequal, then f'(1) doesn't exist (the derivative requires both one-sided derivatives to exist and be equal). However, the problem states f is continuous at x = 1, which is perfectly possible—think of a corner or cusp where the graph doesn't break but changes direction sharply. This situation (continuous but not differentiable) is common at corners, cusps, and points with vertical tangents. Choice E incorrectly claims unequal derivatives imply discontinuity, but corners are continuous. When one-sided derivatives differ, you have non-differentiability, but continuity is a separate question determined by limits and function values.

This question tests the connection between differentiability and continuity of a function at a point. Differentiability at a point implies continuity there because the finite derivative limit ensures the function's limit matches its value at that point. Specifically, for h(x)=|x|, if differentiable at x=0, the difference quotient limit would require continuity, but here we use the theorem in reverse. However, continuity does not imply differentiability, as seen in the absolute value function where left and right slopes differ. A tempting distractor is choice A, which states that if h is continuous at x=0, then it is differentiable there, but this fails because the corner at x=0 means the derivative does not exist despite continuity. A transferable implication strategy is to test one-sided limits of the difference quotient to check differentiability after confirming continuity.

This question tests the connection between differentiability and continuity of a function at a point. Differentiability at a point implies continuity there because the existence of the limit of the difference quotient ensures the function is continuous. Specifically, if r is differentiable at x=a, then by theorem, it must be continuous at x=a. However, continuity does not imply differentiability, as many continuous functions lack derivatives at certain points. A tempting distractor is choice B, which states that if r is continuous at x=a, then it is differentiable there, but this fails because continuity is necessary but not sufficient for differentiability. A transferable implication strategy is to apply the contrapositive: if a function is not continuous, it cannot be differentiable, helping rule out options quickly.

This question tests understanding of boundary behavior for differentiability and continuity. The problem states f is continuous on the closed interval [1,3], which explicitly includes continuity at the endpoints x = 1 and x = 3. However, f is only differentiable on the open interval (1,3), which excludes the endpoints—we know nothing about differentiability at x = 1. This is a common scenario in calculus theorems like the Mean Value Theorem, where continuity extends to endpoints but differentiability might not. Choice A incorrectly assumes differentiability at x = 1, but the given information doesn't guarantee this. When dealing with closed versus open intervals, pay attention to what's guaranteed at endpoints—continuity often extends there, but differentiability might not.

This question tests the relationship between discontinuity types and differentiability. A jump discontinuity at x = 7 means the left and right limits exist but are unequal, so $lim_{x→7}$ f(x) doesn't exist, making f discontinuous at x = 7. Since differentiability requires continuity as a prerequisite, f cannot be differentiable at x = 7 either. The function fails both conditions—it has a break (discontinuous) and certainly no tangent line can exist across a jump. Choice E is absurd because discontinuities prevent derivatives from existing. Remember the hierarchy: for differentiability, you must first have continuity; any discontinuity automatically rules out differentiability.

This question tests the connection between differentiability and continuity. Since f(2) = 5 and $lim_{x→2}$ f(x) = 5, we have continuity at x = 2 because the function value equals the limit. However, we're told that f'(2) does not exist, meaning f is not differentiable at x = 2. This demonstrates that continuity does not guarantee differentiability—a function can be continuous at a point but fail to have a derivative there (like at a corner or cusp). Choice C is impossible because differentiability always implies continuity. When a function is continuous but the derivative doesn't exist, look for geometric features like corners, cusps, or vertical tangents.

This question tests the connection between differentiability and continuity, a key concept in calculus. Differentiability at a point requires that the function has a well-defined tangent line, which implies the function must be continuous there because the limit of the difference quotient must exist and match the function's behavior. However, continuity alone does not guarantee differentiability, as a function can be continuous but have a sharp corner or cusp where no tangent exists. For example, the absolute value function is continuous everywhere but not differentiable at the vertex. A tempting distractor is choice A, which fails because the derivative requires more than just the function being defined; continuity is necessary. To apply this implication strategically, always check continuity first when assessing differentiability, as discontinuity immediately rules out differentiability.

This question tests the connection between differentiability and continuity, a key concept in calculus. Differentiability at a point requires that the function has a well-defined tangent line, which implies the function must be continuous there because the limit of the difference quotient must exist and match the function's behavior. However, continuity alone does not guarantee differentiability, as a function can be continuous but have a sharp corner or cusp where no tangent exists. For example, the absolute value function is continuous everywhere but not differentiable at the vertex. A tempting distractor is choice E, which fails because it claims differentiability without continuity, which contradicts the theorem. To apply this implication strategically, always check continuity first when assessing differentiability, as discontinuity immediately rules out differentiability.

This question tests the connection between differentiability and continuity, a key concept in calculus. Differentiability at a point requires that the function has a well-defined tangent line, which implies the function must be continuous there because the limit of the difference quotient must exist and match the function's behavior. However, continuity alone does not guarantee differentiability, as a function can be continuous but have a sharp corner or cusp where no tangent exists. For example, the absolute value function is continuous everywhere but not differentiable at the vertex. A tempting distractor is choice C, which fails because it suggests differentiability without continuity, which is impossible. To apply this implication strategically, always check continuity first when assessing differentiability, as discontinuity immediately rules out differentiability.

This question tests the connection between differentiability and continuity, a key concept in calculus. Differentiability at a point requires that the function has a well-defined tangent line, which implies the function must be continuous there because the limit of the difference quotient must exist and match the function's behavior. However, continuity alone does not guarantee differentiability, as a function can be continuous but have a sharp corner or cusp where no tangent exists. For example, the absolute value function is continuous everywhere but not differentiable at the vertex. A tempting distractor is choice A, which fails because continuity does not imply differentiability, as stated in the question where the derivative does not exist. To apply this implication strategically, always check continuity first when assessing differentiability, as discontinuity immediately rules out differentiability.