This question assesses the concept of instantaneous rate of change at a point. The average rate of change is the net change in the function divided by the change in the input over an interval. The instantaneous rate of change is the limit of the average rate as the interval approaches zero length around the point. Thus, it is given by the derivative of the function at that point. A tempting distractor is to compute $1 + 1/x^2 = 2$ instead of $1 - 1/x^2$, leading to choice C, but the derivative of $1/x$ is $-1/x^2$. A transferable rate-distinction strategy is to check if the query specifies an interval for average rate or a single point for instantaneous rate.

This problem requires the instantaneous rate of change of f(x) = 1/x at x = 2. Instantaneous rate is found using the derivative at a point, not the average change over an interval. The derivative of 1/x is f'(x) = -1/x². At x = 2: f'(2) = -1/2² = -1/4. The negative sign indicates the function is decreasing at x = 2. Choice B (1/4) might attract students who forget the negative sign when differentiating 1/x. Remember that d/dx(x^(-1)) = -x^(-2) = -1/x².

This problem asks for the average rate of change of q(x) = x² + 1 on [1, 4]. Average rate measures the overall change divided by the interval width, unlike instantaneous rate which uses derivatives. We calculate: [q(4) - q(1)]/(4 - 1) = [(16 + 1) - (1 + 1)]/(4 - 1) = [17 - 2]/3 = 15/3 = 5. This represents the constant slope of the secant line from (1, 2) to (4, 17). Choice B (3) might tempt students who make an arithmetic error or use the wrong interval. For average rates, carefully evaluate the function at both endpoints before applying the formula.

This problem requires the instantaneous rate of change of h(t) = √t at t = 9. The instantaneous rate is found by taking the derivative and evaluating at the specific point, while average rate would use two points. The derivative is h'(t) = 1/(2√t). At t = 9: h'(9) = 1/(2√9) = 1/(2·3) = 1/6. Choice A (1/3) might tempt students who forget the factor of 2 in the denominator of the square root derivative. When differentiating square roots, remember the power rule gives (1/2)t^(-1/2) = 1/(2√t).

This question assesses your understanding of the instantaneous rate of change, found via the derivative at a point. Average rate of change uses $ \frac{p(b) - p(a)}{b - a} $ over an interval, giving a net change ratio. Instantaneous is $ p'(t) = 2t + 3 $, at $ t=2 $ yielding 7, the exact slope there. Average approximates over intervals, but instantaneous is precise at the point. A tempting distractor is choice E, the difference quotient form, but instantaneous requires the derivative, not the average expression. To distinguish, see if it's at a point (derivative for instantaneous) or over an interval (average).

This problem asks for the average rate of change of p(x) = ln x over [1, e²]. Average rate measures the overall change divided by the interval length, unlike instantaneous rate which would use the derivative. We calculate: [p(e²) - p(1)]/(e² - 1) = [ln(e²) - ln(1)]/(e² - 1) = [2 - 0]/(e² - 1) = 2/(e² - 1). Choice B (2) might tempt students who forget to divide by the interval length. For average rate problems with logarithms, remember that $ln(e^n$) = n and ln(1) = 0.

This problem requires calculating average rate of change from tabular data over the interval [4,7]. Using the formula: (k(7) - k(4))/(7 - 4) = (24 - 15)/(7 - 4) = 9/3 = 3. Choice C might confuse students who use the endpoints [1,7] instead of the requested interval [4,7]. Remember that average rate of change measures the overall change per unit input between two specific points, creating a secant line slope rather than the instantaneous tangent line slope.

This question tests understanding of instantaneous rate of change at a point. The instantaneous rate of change of a function at a specific point is given by the derivative at that point, which measures the slope of the tangent line. Average rate of change, in contrast, measures the slope of a secant line between two points using the formula (f(b)-f(a))/(b-a). Since we're given f'(2) = -3, this is the instantaneous rate of change at x = 2. Students might be tempted to choose D (the value 5) by confusing f(2) = 5 (the function value) with the rate of change. Remember: instantaneous rate = derivative value, while average rate = slope of secant line.

This question tests calculation of average rate of change between two points. Average rate of change uses the formula (p(7)-p(3))/(7-3) to find the slope of the secant line. With p(3) = 7 and p(7) = 1, we get (1-7)/(7-3) = -6/4 = -3/2. The negative value indicates the function is decreasing on this interval. Students might choose B by writing (7-1)/(7-3), incorrectly using x-values in the numerator instead of function values. Remember: average rate = (ending y - starting y)/(ending x - starting x), maintaining consistent order in both numerator and denominator.

This question asks for the instantaneous rate of change at t = 0, which requires the derivative. The instantaneous rate gives the slope of the tangent line at that exact point, while average rate would give the slope of a secant line over an interval. For r(t) = 3t⁴ - 2t, we find r'(t) = 12t³ - 2, and r'(0) = 12(0)³ - 2 = 0 - 2 = -2. A student might evaluate r(0) = 0 instead of r'(0), incorrectly choosing answer A. The strategy: instantaneous rate always requires differentiation first, then evaluation at the given point.