Source: Homework Set 0, Problems 1–2
Tags: critical points, second derivative test, local maxima, local minima, Fundamental Theorem of Calculus, FTC, chain rule, Leibniz rule, single-variable calculus review
Difficulty: Introductory review | Prerequisites: Calculus I and II (derivatives, integrals, chain rule)
This homework set is the gateway assignment for MATH 215 (Calculus III). Problems 1 and 2 revisit single-variable calculus to make sure your differentiation and integration toolkit is sharp before the course moves into multivariable territory. If you are rusty on critical-point classification or on differentiating under an integral sign, sort that out here, because both ideas return in higher dimensions (partial derivatives, gradient tests, line integrals). A student who skipped Calc II or took it more than a year ago should work through these problems slowly.
Critical points are where a function's derivative equals zero (or does not exist), and you classify them using the first or second derivative test. The Fundamental Theorem of Calculus lets you differentiate an integral whose limits are functions of x by applying the chain rule to each limit. Both skills are single-variable warm-ups that generalise directly into MATH 215's multivariable framework.
Critical point
A value x = c where f'(c) = 0 or f'(c) does not exist. These are the only candidates for local extrema. In simple terms: the spots where the graph flattens out or has a sharp corner.
Local maximum
A point x = c where f(c) is greater than or equal to f(x) for all x in some open interval around c. Think of it as the top of a hill on the graph.
Local minimum
A point x = c where f(c) is less than or equal to f(x) for all x in some open interval around c. Think of it as the bottom of a valley on the graph.
First Derivative Test
A method for classifying critical points by checking whether f' changes sign (positive to negative = local max, negative to positive = local min, no change = neither). In simple terms: watch which way the slope tilts on either side of the critical point.
Second Derivative Test
If f'(c) = 0 and f''(c) > 0, then c is a local minimum. If f'(c) = 0 and f''(c) < 0, then c is a local maximum. If f''(c) = 0, the test is inconclusive. Think of it as checking the concavity: concave up = min, concave down = max.
Fundamental Theorem of Calculus (FTC)
If F is continuous on [a, b] and f is its derivative, then the integral from a to b of f(x) dx equals F(b) - F(a). Conversely, the derivative of the integral from a to x of f(t) dt equals f(x). In simple terms: differentiation and integration undo each other.
Leibniz integral rule (differentiation under the integral sign)
A rule for differentiating an integral where one or both limits are functions of x. You apply the chain rule to the upper and lower limits separately. In simple terms: to differentiate an integral with variable bounds, evaluate the integrand at each bound and multiply by that bound's derivative.
Compute f'(x) and set it equal to zero. Also check where f'(x) is undefined.
Solve for x to get the critical points.
Classify each critical point using either the first derivative test (sign chart) or the second derivative test (concavity check).
Finding f'(x)
Differentiate term by term: f'(x) = 15x² – 15x⁴ = 15x²(1 – x²) = 15x²(1 – x)(1 + x).
Setting f'(x) = 0
The critical points are x = 0, x = 1, and x = –1.
Classifying with the second derivative test
Compute f''(x) = 30x – 60x³.
At x = 1: f''(1) = 30 – 60 = –30 < 0, so x = 1 is a local maximum.
At x = –1: f''(–1) = –30 + 60 = 30 > 0, so x = –1 is a local minimum.
At x = 0: f''(0) = 0, so the second derivative test is inconclusive.
Resolving the inconclusive case with the first derivative test
Check the sign of f'(x) near x = 0. For small negative x, f'(x) > 0 (increasing). For small positive x, f'(x) > 0 (still increasing). Since the sign does not change, x = 0 is neither a local max nor a local min.
Suppose f(x) is defined as an integral with variable upper and lower limits:
f(x) = integral from g(x) to h(x) of p(u) du
Then by the Leibniz rule (FTC combined with the chain rule):
f'(x) = p(h(x)) · h'(x) – p(g(x)) · g'(x)
The idea: evaluate the integrand at each limit and multiply by that limit's derivative. The upper limit contributes a positive term; the lower limit contributes a negative term.
Step 1: Find f'(x)
Here p(u) = e^(u²), h(x) = x³ (upper limit), g(x) = x (lower limit).
f'(x) = e^((x³)²) · 3x² – e^(x²) · 1 = 3x² e^(x⁶) – e^(x²)
Step 2: Find f''(x)
Differentiate f'(x) using the product rule and chain rule.
For the first term, 3x² e^(x⁶): use the product rule.
d/dx(3x²) = 6x
d/dx(e^(x⁶)) = 6x⁵ e^(x⁶)
Product rule gives: 6x e^(x⁶) + 3x² · 6x⁵ e^(x⁶) = 6x e^(x⁶) + 18x⁷ e^(x⁶)
For the second term, e^(x²): d/dx(e^(x²)) = 2x e^(x²)
So f''(x) = 6x e^(x⁶) + 18x⁷ e^(x⁶) – 2x e^(x²)
Step 3: Evaluate at x = 0
f''(0) = 6(0) e^0 + 18(0)⁷ e^0 – 2(0) e^0 = 0
Every term has a factor of x, so f''(0) = 0.
f'(x) = 0 \quad \text{or } f'(x) \text{ undefined} \implies x \text{ is a critical point}\text{Second Derivative Test: } f''(c) > 0 \implies \text{local min}, \quad f''(c) < 0 \implies \text{local max}\frac{d}{dx}\int_{g(x)}^{h(x)} p(u)\,du = p(h(x))\cdot h'(x) - p(g(x))\cdot g'(x)\int_a^b f(x)\,dx = F(b) - F(a) \quad \text{(Fundamental Theorem of Calculus)}Critical-point analysis is how engineers locate optimal operating conditions: maximising fuel efficiency, minimising material stress, or finding the best antenna angle. Any time you are optimising a quantity that depends on one variable, you are finding and classifying critical points.
Differentiating under the integral sign appears in physics whenever a quantity is defined as an accumulation that depends on a parameter: heat flow through a wall, probability distributions in statistics, or the voltage across a capacitor as its charge changes over time.
Students often assume that f'(c) = 0 guarantees a local max or min. It does not. The point x = 0 in the worked example above is a critical point that is neither, because f' does not change sign there.
When the second derivative test gives f''(c) = 0, students sometimes conclude the point is an inflection point. That is not necessarily true either. You must fall back to the first derivative test or examine higher-order derivatives.
When differentiating an integral with variable limits, students frequently forget the chain-rule factor. The derivative of the upper limit h(x) must multiply the integrand evaluated at h(x). Leaving out h'(x) or g'(x) is the single most common error on this type of problem.
Some students swap the signs and write p(g(x)) g'(x) - p(h(x)) h'(x). The upper limit always gets the positive sign.
⚠️ Critical-point classification is tested on nearly every Calc III midterm, except now with partial derivatives and the second-partials test. Get the single-variable version automatic.
⚠️ Differentiating integrals with variable limits appears again when you work with line integrals and potential functions. If you cannot apply the Leibniz rule quickly, those problems will be painful.
⚠️ The second derivative test being inconclusive is a favourite exam trap. Know what to do when f''(c) = 0.
True or False: If f'(c) = 0, then x = c must be a local maximum or a local minimum.
Fill in the blank: The second derivative test says that if f'(c) = 0 and f''(c) > 0, then c is a local ______.
True or False: To differentiate the integral from x² to x³ of sin(u) du with respect to x, you simply evaluate sin(x³) - sin(x²).
Fill in the blank: When the second derivative test is inconclusive (f''(c) = 0), you should use the ______ test instead.
True or False: The Leibniz integral rule gives the upper limit a negative sign and the lower limit a positive sign.
Answers: 1. False. 2. minimum. 3. False (you must multiply by the derivatives of the limits: 3x² sin(x³) - 2x sin(x²)). 4. first derivative. 5. False (upper limit is positive, lower limit is negative).
Q: Find the critical points of g(x) = x⁴ – 4x² + 1 and classify each.
A: g'(x) = 4x³ – 8x = 4x(x² – 2). Critical points at x = 0, x = √2, x = –√2. g''(x) = 12x² – 8. At x = 0: g''(0) = –8 < 0, so local max. At x = ±√2: g''(±√2) = 24 – 8 = 16 > 0, so both are local minima.
Q: Let h(x) = integral from 1 to x² of cos(t³) dt. Find h'(x).
A: By the Leibniz rule, h'(x) = cos((x²)³) · 2x – 0 = 2x cos(x⁶). The lower limit is constant, so its contribution is zero.
Q: Suppose f'(2) = 0 and f''(2) = 0. What can you conclude about x = 2?
A: The second derivative test is inconclusive. You cannot conclude that x = 2 is a max, a min, or an inflection point without further analysis (e.g. the first derivative test or examining higher derivatives).
Q: Evaluate d/dx of the integral from sin(x) to 1 of e^(t²) dt.
A: Rewrite as the negative of the integral from 1 to sin(x) of e^(t²) dt. Then the derivative is –e^(sin²(x)) · cos(x).
Critical-point classification extends directly to functions of two or more variables in MATH 215. The second derivative test becomes the second-partials test using the Hessian determinant. The intuition is identical: you are checking concavity, but now in multiple directions at once.
The Leibniz integral rule is the single-variable ancestor of differentiation of line integrals and surface integrals with respect to a parameter. It also connects to the concept of potential functions and path independence, both central to the vector calculus portion of the course.
critical point, stationary point, turning point, local extremum, relative maximum, relative minimum, first derivative test, sign chart, second derivative test, concavity test, inflection point, FTC, Fundamental Theorem of Calculus, Leibniz rule, differentiation under the integral sign, variable limits of integration, chain rule for integrals, MATH 215, Calculus III, University of Michigan, single-variable calculus review, optimisation, optimization