Derivative as Slope - Business Calculus

1. At what point $(x, y)$ on the curve $f(x) = x^3 - 6x^2 + 5x$ is the tangent line parallel to the line $y = -4x + 10$ and the $y$-coordinate is negative?
   Answer: (3, -12) The slope of the given line is $-4$. For the tangent line to be parallel, its slope must also be $-4$. The slope of the tangent line to $f(x)$ is given by its derivative, $f'(x) = 3x^2 - 12x + 5$. We set the derivative equal to the desired slope: $3x^2 - 12x + 5 = -4$. This simplifies to the quadratic equation $3x^2 - 12x + 9 = 0$, or $x^2 - 4x + 3 = 0$. Factoring gives $(x-1)(x-3) = 0$, so the x-coordinates are $x=1$ and $x=3$. We find the corresponding y-coordinates: $f(1) = 1^3 - 6(1)^2 + 5(1) = 0$. The point is $(1, 0)$. $f(3) = 3^3 - 6(3)^2 + 5(3) = 27 - 54 + 15 = -12$. The point is $(3, -12)$. The question asks for the point where the y-coordinate is negative, which is $(3, -12)$.
2. The temperature $T(t)$ in a chemical reaction follows $T(t) = 25 + 40e^{-0.2t}$ degrees Celsius, where $t$ is time in minutes. What does the slope of the tangent line at $t = 5$ represent in this context?
   Answer: The temperature is decreasing at approximately 2.95°C per minute at the 5-minute mark The slope of the tangent line at t = 5 is T'(5). First, T'(t) = 40(-0.2)$e^{-0.2t}$ = -$8e^{-0.2t}$. At t = 5: T'(5) = -$8e^{-0.2×5}$ = -$8e^{-1}$ = -8(0.368) ≈ -2.95. The negative sign indicates the temperature is decreasing, and the magnitude 2.95 represents the instantaneous rate of decrease in °C per minute. Choice B has the wrong sign, choice C misinterprets the derivative as a temperature value, and choice D confuses instantaneous rate with average rate.
3. For what quantity $q$ is the instantaneous rate of change of the cost equal to the average cost per item?
   Answer: $20\sqrt{10}$ The instantaneous rate of change of the cost is the derivative of the cost function, also known as the marginal cost, $C'(q)$. $C'(q) = 0.2q + 5$. The average cost per item is $A(q) = \frac{C(q)}{q} = \frac{0.1q^2 + 5q + 400}{q} = 0.1q + 5 + \frac{400}{q}$. We need to find the quantity $q$ where the instantaneous rate of change equals the average cost, so we set $C'(q) = A(q)$: $0.2q + 5 = 0.1q + 5 + \frac{400}{q}$. Subtracting $0.1q$ and 5 from both sides gives $0.1q = \frac{400}{q}$. Multiplying by $q$ gives $0.1q^2 = 400$. Dividing by 0.1 gives $q^2 = 4000$. Taking the square root gives $q = \sqrt{4000} = \sqrt{400 \cdot 10} = 20\sqrt{10}$.
4. What is the x-intercept of the line tangent to the graph of the function $f(x) = x \ln(x)$ at the point where $x=e^2$?
   Answer: $e^2/3$ First, find the point of tangency. At $x_1 = e^2$, the y-coordinate is $y_1 = f(e^2) = e^2 \ln(e^2) = e^2(2) = 2e^2$. The point is $(e^2, 2e^2)$. Next, find the slope of the tangent line by taking the derivative using the product rule: $f'(x) = (1)\ln(x) + x(\frac{1}{x}) = \ln(x) + 1$. The slope at $x=e^2$ is $m = f'(e^2) = \ln(e^2) + 1 = 2 + 1 = 3$. The equation of the tangent line is $y - 2e^2 = 3(x - e^2)$. To find the x-intercept, set $y=0$: $0 - 2e^2 = 3(x - e^2)$. This simplifies to $-2e^2 = 3x - 3e^2$. Adding $3e^2$ to both sides gives $e^2 = 3x$, so $x = e^2/3$.
5. Consider the function $h(x) = \sqrt{x^2 + 16}$. What is the geometric meaning of $h'(4)$?
   Answer: It represents the slope of the tangent line to the curve at the point $(4, 4\sqrt{2})$ h'(4) represents the instantaneous rate of change of h(x) at x = 4, which is geometrically interpreted as the slope of the tangent line to the curve at that point. First, let's find h(4) = √(16 + 16) = √32 = 4√2. The derivative is h'(x) = x/√(x² + 16), so h'(4) = 4/√32 = 4/(4√2) = 1/√2 = √2/2. This is the slope of the tangent line at (4, 4√2). Choice A describes a secant line, choice C describes average rate of change, and choice D describes another secant line.
6. A line that is tangent to the parabola $f(x) = x^2 + 1$ at the point where $x=a$ also passes through the external point $(0, -3)$. What is the value of $a^2$?
   Answer: 4 The derivative of the function is $f'(x) = 2x$. At $x=a$, the slope of the tangent line is $m = f'(a) = 2a$. The point of tangency on the parabola is $(a, f(a))$, which is $(a, a^2+1)$. The equation of the tangent line is $y - (a^2+1) = 2a(x-a)$. Since this line passes through the point $(0, -3)$, we can substitute these values for $x$ and $y$: $-3 - (a^2+1) = 2a(0-a)$. Simplifying the equation gives $-4 - a^2 = -2a^2$. Adding $2a^2$ to both sides gives $a^2 - 4 = 0$, so $a^2 = 4$.
7. Based on the passage, what is the difference in profit between these two production levels?
   Answer: $32 The rate of change of profit is the derivative, $P'(x)$. We need to find where the slope of the tangent line is zero. $P'(x) = 3x^2 - 36x + 96$. Setting $P'(x) = 0$ gives $3(x^2 - 12x + 32) = 0$, which factors as $3(x-4)(x-8) = 0$. The two production levels are $x=4$ and $x=8$. Now we calculate the profit at each level: $P(4) = 4^3 - 18(4^2) + 96(4) + 50 = 64 - 288 + 384 + 50 = 210$. $P(8) = 8^3 - 18(8^2) + 96(8) + 50 = 512 - 1152 + 768 + 50 = 178$. The difference in profit is $|210 - 178| = 32$.
8. The graphs of $f(x) = x^2 + ax + b$ and $g(x) = -x^2 + 1$ are tangent to each other at the point $(1, 0)$. What is the value of $a$?
   Answer: -4 For the two curves to be tangent at a point, they must both pass through that point and have the same slope (the same derivative) at that point. The point $(1, 0)$ is on both curves. For $f(x)$, this means $f(1) = 1^2 + a(1) + b = 0$, so $1+a+b=0$. More importantly, their derivatives must be equal at $x=1$. The derivatives are $f'(x) = 2x + a$ and $g'(x) = -2x$. Setting them equal at $x=1$: $f'(1) = g'(1) \implies 2(1) + a = -2(1)$. This simplifies to $2+a = -2$, which gives $a=-4$.
9. The line $y = 8x + b$ is tangent to the curve $y = 2x^2$. What is the value of the y-intercept, $b$?
   Answer: -8 The slope of the tangent line to the curve $y=2x^2$ is given by its derivative, $y' = 4x$. The given tangent line is $y=8x+b$, which has a slope of 8. To find the point of tangency, we set the derivative equal to the slope: $4x = 8$, which gives $x=2$. Now we find the y-coordinate of this point on the curve: $y = 2(2^2) = 8$. So, the point of tangency is $(2, 8)$. Since this point must lie on the tangent line $y=8x+b$, we substitute its coordinates into the line's equation: $8 = 8(2) + b$. This gives $8 = 16 + b$, so $b = -8$.
10. A function $f(x)$ is differentiable at $x=a$. The tangent line to the graph of $f(x)$ at $x=a$ passes through the origin $(0,0)$. Assuming $a \neq 0$, which equation correctly describes the relationship between $a$, $f(a)$, and the derivative $f'(a)$?
    Answer: $f(a) = a \cdot f'(a)$ The equation of the tangent line to $f(x)$ at $x=a$ can be written in point-slope form as $y - f(a) = f'(a)(x-a)$. The problem states that this line passes through the origin, $(0,0)$. We can substitute $x=0$ and $y=0$ into the equation: $0 - f(a) = f'(a)(0-a)$. This simplifies to $-f(a) = f'(a)(-a)$, which is $-f(a) = -a \cdot f'(a)$. Multiplying both sides by -1 gives the relationship $f(a) = a \cdot f'(a)$.
11. For the function $f(x) = \frac{1}{3}x^3 - \frac{3}{2}x^2 + 5$, the tangent line to its graph is perpendicular to the line $x + 4y = 12$ at two distinct x-values. What is the sum of these x-values?
    Answer: 3 First, find the slope of the given line $x + 4y = 12$. Rearranging to slope-intercept form gives $4y = -x + 12$, so $y = -\frac{1}{4}x + 3$. The slope is $m_1 = -1/4$. The slope of a perpendicular line is the negative reciprocal, $m_2 = -1/(-1/4) = 4$. The slope of the tangent line to $f(x)$ is given by its derivative, $f'(x) = x^2 - 3x$. We set the derivative equal to the required perpendicular slope: $x^2 - 3x = 4$. This gives the quadratic equation $x^2 - 3x - 4 = 0$, which factors as $(x-4)(x+1) = 0$. The two x-values are $x=4$ and $x=-1$. Their sum is $4 + (-1) = 3$.
12. The line $y = 5x - 7$ is tangent to the graph of a differentiable function $f(x)$ at the point $(2, 3)$. What is the value of $\lim_{h \to 0} \frac{f(2+h)-f(2)}{h}$?
    Answer: 5 The expression $\lim_{h \to 0} \frac{f(2+h)-f(2)}{h}$ is the definition of the derivative of $f(x)$ at $x=2$, which is $f'(2)$. The derivative of a function at a point gives the slope of the tangent line at that point. The problem states that the line $y = 5x - 7$ is tangent to the graph of $f(x)$ at $x=2$. The slope of this line is 5. Therefore, $f'(2) = 5$.

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