Average & Instantaneous Rates - Calculus 1

1. For a function $f(x)$, the expression $\frac{f(x) - f(3)}{x-3}$ represents the slope of a secant line through the points $(3, f(3))$ and $(x, f(x))$. What does the quantity $\lim_{x \to 3} \frac{f(x) - f(3)}{x-3}$ represent?
   Answer: The slope of the tangent line to the graph of $f(x)$ at $x=3$. When you see the expression $\frac{f(x) - f(3)}{x-3}$ with a limit as $x$ approaches 3, you're looking at the fundamental definition of a derivative. This is one of the most important limit forms in calculus. The expression $\frac{f(x) - f(3)}{x-3}$ gives the slope of the secant line between points $(3, f(3))$ and $(x, f(x))$. As $x$ gets closer and closer to 3, this secant line approaches the tangent line at the point $(3, f(3))$. The limit $\lim_{x \to 3} \frac{f(x) - f(3)}{x-3}$ captures the slope of that tangent line, which is precisely $f'(3)$ - the derivative of $f$ at $x = 3$. Choice A is incorrect because this limit doesn't represent an average rate of change over an interval. Instead, it's the instantaneous rate of change at a single point. Choice B misunderstands what's happening. The limit doesn't give us the change in slope of secant lines; it gives us what the slope approaches - the tangent line's slope. Choice C confuses slope with y-intercept. This limit tells us nothing about where the tangent line crosses the y-axis. Choice D correctly identifies that this limit represents the slope of the tangent line at $x = 3$. Study tip: Memorize this limit definition of the derivative: $f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}$. When you see this pattern, you're always finding the slope of a tangent line, which equals the derivative at that point.
2. The average rate of change of a function $f(x)$ on the interval $[1, 1+h]$ is $m_h$. The instantaneous rate of change at $x=1$ is $m_0$. If $f''(1) < 0$, what is the relationship between $m_h$ and $m_0$ for small positive values of $h$?
   Answer: $m_h < m_0$ The condition $f''(1) < 0$ means the function is concave down at $x=1$. On a concave down curve, the slope of the secant line from a point $x=1$ to a nearby point $x=1+h$ is less than the slope of the tangent line at $x=1$. The average rate of change $m_h$ is the slope of the secant line, and the instantaneous rate of change $m_0$ is the slope of the tangent line. Therefore, for small positive $h$, $m_h < m_0$.
3. Let $f$ be a differentiable odd function, and assume $f(2) = 6$. What is the average rate of change of $f$ over the interval $[-2, 2]$?
   Answer: 3 The average rate of change of $f$ over $[-2, 2]$ is $\frac{f(2) - f(-2)}{2 - (-2)}$. Since $f$ is an odd function, it satisfies the property $f(-x) = -f(x)$. Therefore, $f(-2) = -f(2) = -6$. Substituting these values into the formula gives $\frac{6 - (-6)}{4} = \frac{12}{4} = 3$.
4. Let $f(x)$ be a function for which $f'(c)$ exists. Let $A_{left}(h) = \frac{f(c) - f(c-h)}{h}$ and $A_{right}(h) = \frac{f(c+h) - f(c)}{h}$ for $h > 0$. Which of the following expressions is equivalent to $f'(c)$?
   Answer: $\lim_{h \to 0} \frac{A_{right}(h) + A_{left}(h)}{2}$ By the definition of the derivative, $f'(c) = \lim_{h \to 0} A_{right}(h)$. Also, by a change of variables (let $k = -h$), we can see that $f'(c) = \lim_{k \to 0} \frac{f(c+k) - f(c)}{k}$. This means $\lim_{h \to 0} \frac{f(c) - f(c-h)}{h} = \lim_{h \to 0} A_{left}(h) = f'(c)$. Since both limits exist and are equal to $f'(c)$, the limit of their average is also $f'(c)$. Let's verify: $\lim_{h \to 0} \frac{A_{right}(h) + A_{left}(h)}{2} = \frac{1}{2} (\lim_{h \to 0} A_{right}(h) + \lim_{h \to 0} A_{left}(h)) = \frac{1}{2}(f'(c) + f'(c)) = f'(c)$. This expression represents the symmetric difference quotient, which is a well-known representation of the derivative. A and C are incorrect; their limits would evaluate to $0$ if the second derivative of $f$ is continuous. They represent the limit of the difference between two quantities that both approach $f'(c)$. D is incorrect; its limit would be $(f'(c))^2$.
5. The average rate of change of a differentiable function $g(x)$ over the interval $[3, 3+h]$ is given by the expression $k(h) = \frac{e^{2h} - 1}{h} + 5(3+h)$. What is the instantaneous rate of change of $g(x)$ at $x=3$?
   Answer: $17$ The instantaneous rate of change at a point is the limit of the average rate of change as the interval width approaches zero. Therefore, we need to find $g'(3) = \lim_{h \to 0} k(h)$. The limit is $\lim_{h \to 0} \left( \frac{e^{2h} - 1}{h} + 5(3+h) \right) = \lim_{h \to 0} \frac{e^{2h} - 1}{h} + \lim_{h \to 0} 5(3+h)$. The second term is straightforward: $\lim_{h \to 0} 5(3+h) = 5(3+0) = 15$. The first term is a standard limit form related to the definition of the derivative. Let $f(x) = e^{2x}$. Then $f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{e^{2h} - e^{0}}{h} = \lim_{h \to 0} \frac{e^{2h} - 1}{h}$. Since $f'(x) = 2e^{2x}$, $f'(0) = 2e^0 = 2$. Thus, the first limit is $2$. Combining the results, the instantaneous rate of change is $2 + 15 = 17$. A is incorrect because it ignores the limit of the first term, assuming it is zero. B is incorrect because it likely stems from misremembering the limit $\lim_{h \to 0} \frac{e^{ah}-1}{h}$ as $1$ instead of $a$, leading to $1+15=16$. D is incorrect because it evaluates the second term at $h=1$ instead of taking the limit as $h \to 0$, leading to $2+5(3+1)=22$.
6. Let $f(x) = 2x^2 + 1$. For what interval $[0, b]$ is the average rate of change of $f(x)$ equal to the instantaneous rate of change at $x=1$?
   Answer: $[0, 2]$ First, find the instantaneous rate of change at $x=1$. The derivative is $f'(x) = 4x$, so $f'(1) = 4(1) = 4$. Next, set up the expression for the average rate of change on $[0, b]$: $\frac{f(b) - f(0)}{b - 0} = \frac{(2b^2 + 1) - (2(0)^2 + 1)}{b} = \frac{2b^2}{b} = 2b$. We set the average rate of change equal to the instantaneous rate of change: $2b = 4$. Solving for $b$ gives $b=2$. Therefore, the interval is $[0, 2]$.
7. The temperature of a substance, $T$, in degrees Celsius, is modeled by the function $T(t) = 100e^{-0.05t}$, where $t$ is the time in minutes. The expression $\frac{T(10) - T(0)}{10}$ represents which of the following?
   Answer: The average rate of cooling from $t=0$ to $t=10$ minutes, in degrees per minute. The expression $\frac{T(b) - T(a)}{b-a}$ is the formula for the average rate of change of the function $T(t)$ over the interval $[a, b]$. In this case, with $a=0$ and $b=10$, the expression represents the average rate of change of temperature with respect to time over the first 10 minutes. The units are degrees per minute, representing a rate of cooling.
8. Let $f(x) = 1/x$. The average rate of change of $f$ over the interval $[1, 4]$ is $-1/4$. The instantaneous rate of change of $f$ at $x=c$ is also $-1/4$. What is the value of $c$?
   Answer: 2 The instantaneous rate of change is the derivative. The derivative of $f(x) = 1/x = x^{-1}$ is $f'(x) = -1x^{-2} = -1/x^2$. We are given that the instantaneous rate of change at $x=c$ is $-1/4$. So, we set $f'(c) = -1/c^2 = -1/4$. This implies $c^2 = 4$, so $c = \pm 2$. Since the original interval is $[1, 4]$, the value must be $c=2$.
9. Let $f(x) = x^2$. Let $AROC_1$ be the average rate of change on $[1, 3]$ and $AROC_2$ be the average rate of change on $[3, 5]$. What is the average rate of change on $[1, 5]$ in terms of $AROC_1$ and $AROC_2$?
   Answer: $\frac{AROC_1 + AROC_2}{2}$ For a quadratic function $f(x) = ax^2+bx+c$, the average rate of change on an interval $[p, q]$ is equal to the instantaneous rate of change at the midpoint, $f'((p+q)/2)$. Here, $f'(x) = 2x$. $AROC_1 = f'((1+3)/2) = f'(2) = 4$. $AROC_2 = f'((3+5)/2) = f'(4) = 8$. The average rate of change on $[1, 5]$ is $f'((1+5)/2) = f'(3) = 6$. Now we check the options. $\frac{AROC_1 + AROC_2}{2} = \frac{4+8}{2} = 6$. This works because the intervals are of equal length. In general, for intervals of equal length $[a, b]$ and $[b, c]$, the AROC on $[a, c]$ is the average of the AROCs on the subintervals.
10. The average rate of change of the function $f(x) = x^3$ on the interval $[2, b]$ is 19. Given that $b>2$, what is the value of $b$?
    Answer: 3 The average rate of change is given by $\frac{f(b) - f(2)}{b-2} = 19$. Substituting $f(x) = x^3$, we get $\frac{b^3 - 2^3}{b-2} = 19$. Factoring the difference of cubes gives $\frac{(b-2)(b^2+2b+4)}{b-2} = 19$. This simplifies to $b^2 + 2b + 4 = 19$, or $b^2 + 2b - 15 = 0$. Factoring the quadratic yields $(b+5)(b-3) = 0$. The solutions are $b = -5$ and $b = 3$. Since the problem states $b>2$, the correct value is $b=3$.
11. Let $f(x)$ be a differentiable function. Which of the following limits represents the instantaneous rate of change of $g(x) = x f(x)$ at $x=a$?
    Answer: $\lim_{h \to 0} \frac{(a+h)f(a+h) - a f(a)}{h}$ The instantaneous rate of change of $g(x)$ at $x=a$ is given by the definition of the derivative: $g'(a) = \lim_{h \to 0} \frac{g(a+h) - g(a)}{h}$. Substituting the expression for $g(x)$, we get $\lim_{h \to 0} \frac{(a+h)f(a+h) - a f(a)}{h}$. This directly matches choice C. The other choices represent parts of the product rule: choice A is $a \cdot f'(a)$, choice D is $f(a)$, and choice B is $f'(a)$. The full derivative by product rule is $g'(a) = f(a) + a f'(a)$.
12. Let $f(x)$ be a differentiable function. If the average rate of change of $f$ on any interval $[x_1, x_2]$ is always equal to $k$ for some constant $k$, which of the following must be true about the instantaneous rate of change, $f'(x)$?
    Answer: $f'(x) = k$ The instantaneous rate of change at a point $c$, $f'(c)$, is the limit of the average rate of change on an interval $[c, c+h]$ as $h \to 0$. If the average rate of change on any interval is always the constant $k$, then $f'(c) = \lim_{h \to 0} (\text{AROC on } [c, c+h]) = \lim_{h \to 0} k = k$. This is true for any point $c$, so $f'(x) = k$. This describes a linear function, $f(x) = kx + b$, whose rate of change is always constant.
13. The expression $\frac{f(x) - f(a)}{x-a}$ represents the slope of a secant line. What does the expression $f(a) + f'(a)(x-a)$ represent in relation to the function $f$ at $x=a$?
    Answer: The equation of the tangent line to the graph of $f$ at the point $x=a$. The expression represents the point-slope form of a line, $y - y_1 = m(x - x_1)$, rewritten as $y = y_1 + m(x-x_1)$. Here, the point $(x_1, y_1)$ is $(a, f(a))$, and the slope $m$ is $f'(a)$, which is the instantaneous rate of change, or the slope of the tangent line at $x=a$. Therefore, the entire expression is the equation of the tangent line to the graph of $f$ at $x=a$.
14. Let a function $f(t)$ represent the amount of water in a tank at time $t$. What is the best interpretation of the expression $\lim_{h \to 0} \frac{f(t_0) - f(t_0 - h)}{h}$?
    Answer: The instantaneous rate at which water is flowing into the tank at time $t_0$. This is a variation of the limit definition of the derivative. Let $k = -h$. As $h \to 0$, $k \to 0$. The expression becomes $\lim_{k \to 0} \frac{f(t_0) - f(t_0 + k)}{-k} = \lim_{k \to 0} \frac{-(f(t_0+k) - f(t_0))}{-k} = \lim_{k \to 0} \frac{f(t_0+k) - f(t_0)}{k}$. This is the standard definition of $f'(t_0)$, the instantaneous rate of change of the amount of water at time $t_0$. A positive rate signifies water flowing in.
15. The position of a particle is given by $s(t) = t^3 - 6t^2 + 9t$ for $t \ge 0$. The particle's velocity is $v(t) = s'(t)$. What is the average rate of change of the velocity over the time interval during which the particle's velocity is decreasing?
    Answer: -6 First, find the velocity function: $v(t) = s'(t) = 3t^2 - 12t + 9$. To find where the velocity is decreasing, we need to find where its derivative, the acceleration $a(t)$, is negative. $a(t) = v'(t) = 6t - 12$. Setting $a(t) < 0$ gives $6t - 12 < 0$, which means $t < 2$. The interval is $[0, 2]$. Now, we find the average rate of change of $v(t)$ on $[0, 2]$: $\frac{v(2) - v(0)}{2 - 0} = \frac{(3(2^2) - 12(2) + 9) - (9)}{2} = \frac{(12 - 24 + 9) - 9}{2} = \frac{-3 - 9}{2} = \frac{-12}{2} = -6$.
16. If $f'(2) = 5$, which of the following statements must be true?
    Answer: $\lim_{h \to 0} \frac{f(2+h) - f(2)}{h} = 5$ When you see $f'(2) = 5$, you're looking at the derivative of function $f$ evaluated at $x = 2$. This is fundamentally about understanding what a derivative actually represents. The derivative $f'(2) = 5$ tells us the instantaneous rate of change of $f$ at the point $x = 2$. By definition, the derivative is the limit of the difference quotient, so $f'(2) = \lim_{h \to 0} \frac{f(2+h) - f(2)}{h} = 5$. This makes choice D correct—it's literally the definition of what $f'(2)$ means. Choice A is wrong because the average rate of change over $[1,3]$ would be $\frac{f(3) - f(1)}{3 - 1}$, which could be any value. The instantaneous rate at one point doesn't determine the average rate over an interval. Choice B confuses a single point with an entire interval. Knowing $f'(2) = 5 > 0$ tells us $f$ is increasing at $x = 2$, but we need $f'(x) > 0$ for all $x$ in $(1,3)$ to conclude $f$ is increasing on that whole interval. Choice C mixes up the function value with its derivative. $f'(2) = 5$ gives us the slope at $x = 2$, not the actual height $f(2)$ of the function there. Remember this key distinction: the derivative at a point gives you the instantaneous rate of change (slope of the tangent line) at that specific point, which is always defined as a limit of difference quotients. Don't confuse derivatives with function values or behavior over intervals.
17. For the function $f(x) = |x - 3|$, which of the following best describes the instantaneous rate of change at $x=3$?
    Answer: The instantaneous rate of change does not exist. The instantaneous rate of change is the derivative, which is defined as a limit. For $f(x) = |x - 3|$ at $x=3$, we must check if $\lim_{h \to 0} \frac{|(3+h)-3| - |3-3|}{h} = \lim_{h \to 0} \frac{|h|}{h}$ exists. The left-hand limit (as $h \to 0^-$) is $\lim_{h \to 0^-} \frac{-h}{h} = -1$. The right-hand limit (as $h \to 0^+$) is $\lim_{h \to 0^+} \frac{h}{h} = 1$. Since the left and right limits are not equal, the overall limit does not exist, and thus the instantaneous rate of change does not exist at $x=3$.
18. The limit $L = \lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}$ can be interpreted as the ratio of the instantaneous rates of change of two functions $f(x)=x^3$ and $g(x)=x^2$ at $x=2$. What is the value of $L$?
    Answer: 3 The limit can be solved by factoring the numerator and denominator: $L = \lim_{x \to 2} \frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)} = \lim_{x \to 2} \frac{x^2+2x+4}{x+2}$. Substituting $x=2$ gives $\frac{2^2+2(2)+4}{2+2} = \frac{4+4+4}{4} = \frac{12}{4} = 3$. Alternatively, recognizing the structure $\lim_{x \to a} \frac{f(x)-f(a)}{g(x)-g(a)}$, which is equivalent to $\frac{f'(a)}{g'(a)}$ (by L'Hôpital's Rule or algebraic manipulation). Here $f(x)=x^3$ and $g(x)=x^2$. Then $f'(x)=3x^2$ and $g'(x)=2x$. At $a=2$, $f'(2)=12$ and $g'(2)=4$. The ratio is $12/4=3$.
19. The average rate of change of a strictly increasing function $f(x)$ on the interval $[2, 5]$ is $1/3$. Let $g(x)$ be the inverse function of $f(x)$. What is the average rate of change of $g(x)$ on the interval $[f(2), f(5)]$?
    Answer: 3 Let $a = f(2)$ and $b = f(5)$. The average rate of change of $g(x)$ on $[a, b]$ is $\frac{g(b) - g(a)}{b - a}$. Since $g$ is the inverse of $f$, $g(b) = g(f(5)) = 5$ and $g(a) = g(f(2)) = 2$. The expression becomes $\frac{5 - 2}{f(5) - f(2)}$. We are given that the average rate of change of $f$ on $[2, 5]$ is $\frac{f(5) - f(2)}{5 - 2} = \frac{1}{3}$. This means $\frac{f(5) - f(2)}{3} = \frac{1}{3}$, so $f(5) - f(2) = 1$. Substituting this into the expression for the average rate of $g$ gives $\frac{3}{1} = 3$.
20. Let $g(x) = \frac{1}{x+1}$. Find the value $c$ in the interval $(0, 2)$ where the instantaneous rate of change of $g(x)$ at $x=c$ is equal to the average rate of change of $g(x)$ over the interval $[0, 2]$.
    Answer: $c = \sqrt{3} - 1$ First, calculate the average rate of change (AROC) of $g(x)$ over $[0, 2]$: $\frac{g(2) - g(0)}{2 - 0} = \frac{1/3 - 1}{2} = \frac{-2/3}{2} = -\frac{1}{3}$. Next, find the instantaneous rate of change (IROC) by finding the derivative: $g'(x) = -\frac{1}{(x+1)^2}$. Set the IROC at $c$ equal to the AROC: $-\frac{1}{(c+1)^2} = -\frac{1}{3}$. This simplifies to $(c+1)^2 = 3$, so $c+1 = \pm\sqrt{3}$, which gives $c = -1 \pm \sqrt{3}$. Since $c$ must be in the interval $(0, 2)$, we choose the positive root, $c = -1 + \sqrt{3} \approx 0.732$.

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