Differentiation Applications - IB Mathematics Analysis and Approaches

1. Find the maximum value of the function $f(x) = 2x^3 - 9x^2 + 12x$ on the interval $[0, 3]$.
   Answer: 9 To find the absolute maximum value on a closed interval, we must evaluate the function at its critical points within the interval and at the endpoints of the interval. First, find the critical points by setting the derivative to zero: $f'(x) = 6x^2 - 18x + 12$. Set $f'(x) = 0$: $6(x^2 - 3x + 2) = 0 \implies 6(x-1)(x-2) = 0$. The critical points are $x=1$ and $x=2$, both of which are in the interval $[0, 3]$. Now, evaluate the function at the critical points and the endpoints: $f(0) = 0$ $f(1) = 2 - 9 + 12 = 5$ $f(2) = 2(8) - 9(4) + 12(2) = 16 - 36 + 24 = 4$ $f(3) = 2(27) - 9(9) + 12(3) = 54 - 81 + 36 = 9$ Comparing the values {0, 5, 4, 9}, the maximum value is 9.
2. The first derivative of a function $f$ is given by $f'(x) = \frac{x-4}{x^2+1}$. For which interval is $f(x)$ increasing?
   Answer: $x > 4$ When you encounter a question about where a function is increasing or decreasing, you need to analyze the sign of the first derivative. A function is increasing where $f'(x) > 0$ and decreasing where $f'(x) < 0$. Given $f'(x) = \frac{x-4}{x^2+1}$, you need to determine where this expression is positive. Since $x^2+1$ is always positive (it's never zero and always greater than 1), the sign of $f'(x)$ depends entirely on the numerator $x-4$. The numerator $x-4$ is positive when $x > 4$ and negative when $x < 4$. Therefore, $f'(x) > 0$ when $x > 4$, making the function increasing on this interval. This confirms answer D is correct. Looking at the wrong answers: A) $x < 4$ represents where $f'(x) < 0$, so this is actually where the function is decreasing, not increasing. B) $-1 < x < 1$ is a subset of $x < 4$, so $f'(x)$ is still negative here, meaning the function is decreasing. C) $x < -1$ or $x > 1$ incorrectly suggests the function behavior changes at $x = \pm 1$, but the critical point where $f'(x) = 0$ occurs at $x = 4$, not at $\pm 1$. Remember: always find where the derivative equals zero to identify critical points, then test the sign of the derivative on either side of these points to determine intervals of increase and decrease.
3. The position of a particle moving along a straight line is given by $s(t) = t^3 - 12t^2 + 36t$ for $t \ge 0$, where $t$ is in seconds and $s$ is in meters. At what time $t$ does the particle first momentarily stop?
   Answer: 2 s When you encounter a particle motion problem asking when the particle "momentarily stops," you need to find when the velocity equals zero. The velocity is the first derivative of the position function. Given $s(t) = t^3 - 12t^2 + 36t$, let's find the velocity by taking the derivative: $v(t) = s'(t) = 3t^2 - 24t + 36$ To find when the particle stops, set the velocity equal to zero: $3t^2 - 24t + 36 = 0$ Divide everything by 3 to simplify: $t^2 - 8t + 12 = 0$ Factor this quadratic equation: $(t - 6)(t - 2) = 0$ This gives us $t = 2$ or $t = 6$. Since the question asks for the first time the particle stops, the answer is $t = 2$ seconds, which is choice D. Let's examine the wrong answers: Choice A (0 s) occurs because at $t = 0$, the velocity is $v(0) = 36 \neq 0$, so the particle is moving initially. Choice B (6 s) is when the particle stops for the second time, not the first. Choice C (4 s) might tempt you if you made an error in factoring or solving the quadratic—at $t = 4$, we get $v(4) = -12 \neq 0$. Remember: "momentarily stops" means velocity equals zero, not position equals zero. Always take the derivative of position to get velocity, then solve $v(t) = 0$. Watch for multiple solutions and choose the first positive time when asked.
4. Find the equation of the normal to the curve $y = \ln(x^2+1)$ at the point where $x=1$.
   Answer: $y = -x + \ln(2) + 1$ First, find the point of interest. When $x=1$, $y = \ln(1^2+1) = \ln(2)$. The point is $(1, \ln(2))$. Next, find the gradient of the tangent by differentiating $y$. Using the chain rule, $\frac{dy}{dx} = \frac{1}{x^2+1} \cdot 2x = \frac{2x}{x^2+1}$. At $x=1$, the gradient of the tangent is $m_T = \frac{2(1)}{1^2+1} = 1$. The gradient of the normal is the negative reciprocal of the tangent's gradient, so $m_N = -1/1 = -1$. The equation of the normal is given by $y - y_1 = m_N(x - x_1)$. Substituting the point and gradient gives $y - \ln(2) = -1(x - 1)$, which simplifies to $y - \ln(2) = -x + 1$, or $y = -x + 1 + \ln(2)$.
5. The function $f$ is defined by $f(x) = x^3 - 6x^2 + 5$. Find the set of values of $x$ for which $f(x)$ is a decreasing function.
   Answer: $0 < x < 4$ A function is decreasing when its first derivative is negative. First, find the derivative of $f(x)$: $f'(x) = 3x^2 - 12x$. We need to find where $f'(x) < 0$. We solve $3x^2 - 12x < 0$, which is $3x(x - 4) < 0$. The roots of the quadratic $3x(x-4)=0$ are $x=0$ and $x=4$. Since the quadratic is a parabola opening upwards, it will be negative between its roots. Therefore, the function is decreasing for $0 < x < 4$.
6. A rectangular field is to be enclosed by a fence. One side of the field is along a straight river and does not need fencing. The total length of the fence used for the other three sides is 200 metres. Find the maximum possible area of the field.
   Answer: 5000 m² Let the side parallel to the river have length $y$ and the other two sides have length $x$. The total length of the fence is $2x + y = 200$, so $y = 200 - 2x$. The area of the field is $A = xy = x(200 - 2x) = 200x - 2x^2$. To maximize the area, we find the derivative with respect to $x$ and set it to zero: $A'(x) = 200 - 4x$. Setting $A'(x) = 0$ gives $200 - 4x = 0$, which solves to $x = 50$. The corresponding length of the other side is $y = 200 - 2(50) = 100$. The maximum area is $A = 50 \times 100 = 5000$ m². The second derivative $A''(x) = -4 < 0$, confirming it is a maximum.
7. The function $f(x)$ is twice differentiable. If $f'(x) > 0$ and $f''(x) < 0$ for all $x$ in the interval $(a, b)$, which statement best describes the graph of $f(x)$ on this interval?
   Answer: Increasing and concave down The sign of the first derivative determines whether the function is increasing or decreasing. Since $f'(x) > 0$, the function $f(x)$ is increasing. The sign of the second derivative determines the concavity of the function. Since $f''(x) < 0$, the function $f(x)$ is concave down. Therefore, the graph of $f(x)$ is increasing and concave down on the interval $(a, b)$.
8. A rectangle has a perimeter of 24 cm. What is the maximum possible area of this rectangle?
   Answer: 36 cm² Let the length and width of the rectangle be $l$ and $w$ respectively. The perimeter is $2l + 2w = 24$, which simplifies to $l + w = 12$. We can express $w$ as $w = 12 - l$. The area is $A = l \times w = l(12 - l) = 12l - l^2$. To find the maximum area, we take the derivative of $A$ with respect to $l$ and set it to zero: $A'(l) = 12 - 2l$. Setting $A'(l) = 0$ gives $12 - 2l = 0$, so $l = 6$. When $l = 6$, $w = 12 - 6 = 6$. The rectangle with maximum area is a square. The maximum area is $A = 6 \times 6 = 36$ cm².
9. The function $f(x) = x^3 + px^2 + qx + r$ has a local maximum at $x=0$ and a point of inflection at $x=1$. Find the value of $p$.
   Answer: -3 First, find the first and second derivatives of $f(x)$. $f'(x) = 3x^2 + 2px + q$ and $f''(x) = 6x + 2p$. A local maximum at $x=0$ means $f'(0) = 0$. Substituting $x=0$ into $f'(x)$ gives $3(0)^2 + 2p(0) + q = 0$, so $q=0$. A point of inflection at $x=1$ means $f''(1) = 0$. Substituting $x=1$ into $f''(x)$ gives $6(1) + 2p = 0$, so $6 + 2p = 0$, which solves to $2p = -6$, and $p = -3$. We can check the local maximum condition using the second derivative: $f''(0) = 6(0) + 2p = 2p$. For a maximum, $f''(0) < 0$, so $2p < 0$, which means $p < 0$. Our result $p=-3$ is consistent with this.
10. Find the coordinates of the point of inflection of the curve $f(x) = xe^{-x}$.
    Answer: $(2, 2e^{-2})$ A point of inflection occurs where the second derivative changes sign, which typically happens where the second derivative is zero. First, find the first derivative using the product rule: $f'(x) = 1 \cdot e^{-x} + x \cdot (-e^{-x}) = e^{-x}(1-x)$. Now find the second derivative, again using the product rule: $f''(x) = -e^{-x}(1-x) + e^{-x}(-1) = e^{-x}(-1+x-1) = e^{-x}(x-2)$. Set $f''(x) = 0$. Since $e^{-x}$ is never zero, we must have $x-2=0$, which gives $x=2$. To find the y-coordinate, substitute $x=2$ back into the original function: $f(2) = 2e^{-2}$. The coordinates of the point of inflection are $(2, 2e^{-2})$. We can verify that the sign of $f''(x)$ changes at $x=2$.
11. The second derivative of a function is given by $f''(x) = (x-1)(x-3)^2$. How many points of inflection does the graph of $y=f(x)$ have?
    Answer: 1 A point of inflection occurs where the second derivative $f''(x)$ is zero and changes sign. The possible points of inflection are where $f''(x) = 0$, which occurs at $x=1$ and $x=3$. We must test the sign of $f''(x)$ around these points. The term $(x-3)^2$ is always non-negative. Therefore, the sign of $f''(x)$ is determined by the sign of the term $(x-1)$. For $x < 1$, $x-1$ is negative, so $f''(x) < 0$. For $x > 1$ (but $x \neq 3$), $x-1$ is positive, so $f''(x) > 0$. Since the sign of $f''(x)$ changes at $x=1$, it is a point of inflection. Around $x=3$, for example at $x=2.9$ and $x=3.1$, the term $x-1$ is positive, and $(x-3)^2$ is positive. So the sign of $f''(x)$ does not change at $x=3$. Thus, $x=3$ is not a point of inflection. There is only one point of inflection.
12. Find the x-coordinates of the stationary points of the function $f(x) = x^4 - 4x^3 + 10$.
    Answer: $x=0$ and $x=3$ Stationary points occur where the first derivative of the function is equal to zero. First, find the derivative of $f(x)$: $f'(x) = 4x^3 - 12x^2$. Now, set the derivative to zero and solve for $x$: $4x^3 - 12x^2 = 0$. We can factor out $4x^2$: $4x^2(x - 3) = 0$. The solutions are found by setting each factor to zero: $4x^2 = 0 \implies x = 0$ and $x - 3 = 0 \implies x = 3$. Thus, the x-coordinates of the stationary points are 0 and 3.
13. A function $f(x)$ has derivatives $f'(x)$ and $f''(x)$. It is known that $f'(3) = 0$ and $f''(3) = -4$. Which statement must be true about the graph of $y=f(x)$ at $x=3$?
    Answer: It has a local maximum. The conditions given are for the second derivative test for local extrema. The first condition, $f'(3) = 0$, indicates that there is a stationary point at $x=3$. The second condition, $f''(3) = -4 < 0$, indicates that the curve is concave down at this stationary point. A stationary point on a concave down section of a curve is a local maximum.
14. Find the value of $c$ for which the line $y = 4x + c$ is tangent to the curve $y = x^2$.
    Answer: -4 For the line to be tangent to the curve, their gradients must be equal at the point of tangency. The gradient of the curve $y = x^2$ is given by its derivative, $y' = 2x$. The gradient of the line $y = 4x + c$ is 4. Setting the gradients equal gives $2x = 4$, so $x = 2$. This is the x-coordinate of the point of tangency. The y-coordinate is found by substituting $x=2$ into the curve's equation: $y = 2^2 = 4$. So the point of tangency is $(2, 4)$. This point must also lie on the tangent line. Substituting $(2, 4)$ into $y = 4x + c$ gives $4 = 4(2) + c$, which simplifies to $4 = 8 + c$, so $c = -4$.
15. A cylindrical can with a closed top and bottom is to be made to hold a volume of $1000 \text{ cm}^3$. The cost of the material for the top and bottom is twice the cost per unit area of the material for the side. Find the radius $r$ that minimizes the cost of the can.
    Answer: $\sqrt[3]{\frac{250}{\pi}}$ Let $r$ be the radius and $h$ be the height. The volume is $V = \pi r^2 h = 1000$, so $h = \frac{1000}{\pi r^2}$. Let the cost of the side material be $k$ per cm². The cost of the top/bottom material is (2k). The area of the side is $A_s = 2\pi rh$ and the area of the top and bottom is $A_{tb} = 2\pi r^2$. The total cost is $C = k A_s + 2k A_{tb} = k(2\pi rh) + 2k(2\pi r^2) = 2k\pi(rh + 2r^2)$. Substitute $h$: $C(r) = 2k\pi(r \frac{1000}{\pi r^2} + 2r^2) = 2k\pi(\frac{1000}{\pi r} + 2r^2) = \frac{2000k}{r} + 4k\pi r^2$. To minimize cost, find the derivative and set to zero: $C'(r) = -\frac{2000k}{r^2} + 8k\pi r$. Setting $C'(r)=0$ gives $8k\pi r = \frac{2000k}{r^2}$. The $k$ cancels out. $8\pi r^3 = 2000 \implies r^3 = \frac{2000}{8\pi} = \frac{250}{\pi}$. Thus, $r = \sqrt[3]{\frac{250}{\pi}}$.
16. A company's profit, $P$, in thousands of dollars, from selling $x$ hundred items is given by $P(x) = -x^3 + 9x^2 - 15x$. For which number of items sold is the profit maximized?
    Answer: 500 To find the maximum profit, we need to find the stationary points of the profit function $P(x)$ by setting its derivative to zero. $P'(x) = -3x^2 + 18x - 15$. Set $P'(x) = 0$: $-3(x^2 - 6x + 5) = 0 \implies -3(x-1)(x-5) = 0$. The stationary points are at $x=1$ and $x=5$. To determine which is a maximum, we use the second derivative test: $P''(x) = -6x + 18$. At $x=1$, $P''(1) = -6(1) + 18 = 12 > 0$, which indicates a local minimum. At $x=5$, $P''(5) = -6(5) + 18 = -30 + 18 = -12 < 0$, which indicates a local maximum. Since $x$ represents hundreds of items, $x=5$ corresponds to 500 items.
17. The cost $C$ of producing $x$ items is given by $C(x) = 0.1x^3 - 6x^2 + 150x + 200$, for $x > 0$. Find the number of items that should be produced to minimize the marginal cost.
    Answer: 20 The marginal cost, $M(x)$, is the derivative of the cost function, $C(x)$. So, $M(x) = C'(x) = 0.3x^2 - 12x + 150$. To minimize the marginal cost, we need to find the derivative of $M(x)$ and set it to zero. $M'(x) = 0.6x - 12$. Setting $M'(x) = 0$ gives $0.6x - 12 = 0$, which solves to $x = 12 / 0.6 = 20$. To confirm this is a minimum, we check the second derivative: $M''(x) = 0.6 > 0$, which confirms a minimum. So, 20 items should be produced to minimize the marginal cost.
18. The graph of a function $y=f(x)$ is concave down for all $x<3$ and concave up for all $x>3$. Which of the following is a possible expression for $f''(x)$?
    Answer: $f''(x) = x-3$ The concavity of a function is determined by the sign of its second derivative. Concave down corresponds to $f''(x) < 0$, and concave up corresponds to $f''(x) > 0$. We are given that the graph is concave down for $x<3$ (so $f''(x)<0$ for $x<3$) and concave up for $x>3$ (so $f''(x)>0$ for $x>3$). We need to check which of the given expressions for $f''(x)$ satisfies these conditions. A. $f''(x) = 3-x$: If $x<3$, $3-x>0$ (concave up). Incorrect. B. $f''(x) = (x-3)^2$: This expression is non-negative for all $x$, so it doesn't describe a change from concave down to up. Incorrect. C. $f''(x) = x-3$: If $x<3$, $x-3<0$ (concave down). If $x>3$, $x-3>0$ (concave up). This matches the description. Correct. D. $f''(x) = 9-x^2$: This is a downward-opening parabola with roots at $x=\pm3$. It is negative for $x>3$ and for $x<-3$. Incorrect.
19. The derivative of a function $f$ is given by $f'(x) = x(x-2)^2$. At which value of $x$ does $f(x)$ have a local minimum?
    Answer: $x=0$ Local extrema occur at stationary points, where $f'(x) = 0$. Setting $x(x-2)^2 = 0$ gives stationary points at $x=0$ and $x=2$. To determine the nature of these points, we use the first derivative test. We check the sign of $f'(x)$ in the intervals around the stationary points. For $x < 0$ (e.g., $x=-1$), $f'(-1) = (-1)(-3)^2 = -9 < 0$. For $0 < x < 2$ (e.g., $x=1$), $f'(1) = (1)(-1)^2 = 1 > 0$. Since the sign of $f'(x)$ changes from negative to positive at $x=0$, there is a local minimum at $x=0$. For $x > 2$ (e.g., $x=3$), $f'(3) = (3)(1)^2 = 3 > 0$. The sign does not change at $x=2$, so it is a stationary point of inflection.

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