Corrected exercises: concavity and curve sketching (MATH 203)

a) $f'(x) = 4x^3 + 6x^2 - 24x$ and $f''(x) = 12x^2 + 12x - 24 = 12(x^2 + x - 2) = 12(x + 2)(x - 1)$. Factor BEFORE looking at signs: a product of two linear factors has the sign $+$ outside its roots and $-$ between them. So $f'' > 0$ on $(-\infty, -2)$ and $(1, \infty)$, where $f$ is concave up, and $f'' < 0$ on $(-2, 1)$, where it is concave down. The sign changes at $-2$ and at $1$, both in the domain, and a polynomial has a tangent line everywhere: two points of inflection, $f(-2) = 16 - 16 - 48 + 3 = -45$ and $f(1) = 1 + 2 - 12 + 3 = -6$. A point of inflection is a POINT: writing only $x = -2$ and $x = 1$ loses half a mark on most MATH 203 finals.

b) Chain rule with inner function $x - 2$: $g'(x) = 4(x - 2)^3 + 3$ and $g''(x) = 12(x - 2)^2$, so $g''(2) = 0$. But $(x - 2)^2 \ge 0$ on both sides of $2$: $g''$ does NOT change sign, and $g$ is concave up on $(-\infty, 2)$ and on $(2, \infty)$, hence on all of $\mathbb{R}$. The candidate is rejected: no inflection point. This is why the question says not to expand: $(x - 2)^4 = x^4 - 8x^3 + 24x^2 - 32x + 16$ gives $g''(x) = 12x^2 - 48x + 48$, and the student who expands must now FACTOR it back into $12(x - 2)^2$ to read its sign. The factored form was there from the start; expanding it costs time on an exam and often a sign.

c) Differentiate term by term with the power rule: $h'(x) = \frac{7}{3}x^{4/3} - \frac{7}{3}x^{-2/3}$ and $h''(x) = \frac{28}{9}x^{1/3} + \frac{14}{9}x^{-5/3}$. Now the algebraic gesture of the chapter: factor out the SMALLEST power, $x^{-5/3}$. Since $x^{1/3} = x^{-5/3} \cdot x^{2}$, $h''(x) = \frac{14}{9}x^{-5/3}(2x^2 + 1) = \frac{14(2x^2 + 1)}{9x^{5/3}}$. The factor $2x^2 + 1 \ge 1$ never vanishes, so $h''(x) = 0$ has NO solution. Yet $x^{5/3} = \left(\sqrt[3]{x}\right)^5$ has the sign of $x$: $h'' < 0$ for $x < 0$ (concave down) and $h'' > 0$ for $x > 0$ (concave up). The candidate is $x = 0$, where $h''$ does not exist but $h(0) = 0$ does. The tangent line: $h'(x) = \frac{7(x^2 - 1)}{3x^{2/3}}$, and as $x \to 0$ the numerator tends to $-7$ while $3x^{2/3} \to 0^+$ from both sides, so $h'(x) \to -\infty$: the graph has a VERTICAL tangent at the origin, which the figure shows. With a tangent line and a change of concavity, $(0, 0)$ is a point of inflection. Note that $h''$ written as the SUM $\frac{28}{9}x^{1/3} + \frac{14}{9}x^{-5/3}$ hides everything: its sign is unreadable until the smallest power is out.

d) Not sufficient: in b), $g''(2) = 0$ and $(2, g(2))$ is not a point of inflection, because the sign of $g''$ does not change. Not necessary: in c), $h''$ never vanishes and $(0, 0)$ is a point of inflection, because the candidate there is a point where $h''$ does not exist. The correct rule: list the zeros of $f''$ AND the points of the domain where $f''$ is undefined, then keep a candidate only if $f''$ changes sign there and the graph has a tangent line.

a) $f'(x) = 3x^2 - 12x - 15 = 3(x^2 - 4x - 5) = 3(x + 1)(x - 5)$, so the critical numbers are $-1$ and $5$. $f''(x) = 6x - 12$. At $-1$: $f''(-1) = -18 < 0$, local maximum $f(-1) = -1 - 6 + 15 + 4 = 12$. At $5$: $f''(5) = 18 > 0$, local minimum $f(5) = 125 - 150 - 75 + 4 = -96$. The test is faster than a sign table here because $f''$ is a one-line computation and never vanishes at a critical number. The classic slip is the sign of $f''$ itself: $f''(-1)$ NEGATIVE means concave down, the shape of a cap, and a cap on a horizontal tangent is a maximum.

b) $g'(x) = 1 - 2\sin x = 0 \iff \sin x = \frac{1}{2}$, which on $(0, 2\pi)$ gives $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$ (the second one comes from $\sin(\pi - \theta) = \sin\theta$, and forgetting it is the usual loss). $g''(x) = -2\cos x$. At $\frac{\pi}{6}$: $g'' = -2 \cdot \frac{\sqrt 3}{2} = -\sqrt 3 < 0$, local maximum $g\left(\frac{\pi}{6}\right) = \frac{\pi}{6} + 2 \cdot \frac{\sqrt 3}{2} = \frac{\pi}{6} + \sqrt 3 \approx 2.2556$. At $\frac{5\pi}{6}$: $g'' = -2 \cdot \left(-\frac{\sqrt 3}{2}\right) = \sqrt 3 > 0$, local minimum $g\left(\frac{5\pi}{6}\right) = \frac{5\pi}{6} - \sqrt 3 \approx 0.8859$. Calculator in RADIAN mode: in degree mode, $\cos\frac{\pi}{6}$ is computed as the cosine of $0.52$ degrees and every decimal is wrong. The figure agrees: a small bump then a small dip, and the line $y = x$ wins after that.

c) $p'(x) = 4x^3 + 12x^2 = 4x^2(x + 3)$: factor out the common $4x^2$ instead of dividing by $x$, which would lose the root $0$. Critical numbers $0$ and $-3$. $p''(x) = 12x^2 + 24x = 12x(x + 2)$. At $-3$: $p''(-3) = 36 > 0$, local minimum $p(-3) = 81 - 108 = -27$. At $0$: $p''(0) = 0$, the test is SILENT. It does not say no extremum; it says nothing. The First Derivative Test decides: $4x^2 \ge 0$, so near $0$ the sign of $p'$ is the sign of $x + 3$, positive on both sides of $0$. No sign change, no extremum. Moreover $p'' = 12x(x + 2)$ changes sign at $0$ (from $-$ to $+$): the origin is a point of inflection where the tangent line is HORIZONTAL, a flat step in the graph.

d) $k'(x) = 4x^3 - 12x^2 + 12x - 4$, so $k'(1) = 4 - 12 + 12 - 4 = 0$, and $k''(x) = 12x^2 - 24x + 12$, so $k''(1) = 0$: the test is silent again. Now look at the coefficients $1, -4, 6, -4$: they are those of $(x - 1)^4 = x^4 - 4x^3 + 6x^2 - 4x + 1$. So $k(x) = (x - 1)^4 + 2$. Since $(x - 1)^4 \ge 0$ with equality only at $x = 1$, $k(x) \ge 2 = k(1)$ for every $x$: $k$ has an ABSOLUTE minimum $2$ at $x = 1$. Compare with c): at both points $f' = f'' = 0$, and the conclusions are opposite (no extremum in c), a minimum in d)). That is exactly what the word fails means. The algebra did the work: recognizing the binomial replaced a whole sign table, and $k'(x) = 4(x - 1)^3$ is then read at a glance.

a) $x^2 + 27 \ge 27 > 0$, so the domain is $\mathbb{R}$ and there is no vertical asymptote. $f(0) = -\frac{9}{27} = -\frac{1}{3}$; $f(x) = 0 \iff x^2 = 9 \iff x = \pm 3$. Only $x^2$ appears, so $f(-x) = f(x)$: $f$ is even. Add and subtract $36$ in the numerator: $\frac{x^2 - 9}{x^2 + 27} = \frac{(x^2 + 27) - 36}{x^2 + 27} = 1 - \frac{36}{x^2 + 27}$. As $x \to \pm\infty$, $\frac{36}{x^2 + 27} \to 0$, so $y = 1$ is the horizontal asymptote at both ends; and since $\frac{36}{x^2 + 27} > 0$, $f(x) < 1$ everywhere: the graph stays BELOW its asymptote and never reaches it.

b) Quotient rule: $f'(x) = \frac{2x(x^2 + 27) - (x^2 - 9)(2x)}{(x^2 + 27)^2}$. Factor the common $2x$ out of the numerator before expanding anything: $2x\left[(x^2 + 27) - (x^2 - 9)\right] = 2x \cdot 36 = 72x$. So $f'(x) = \frac{72x}{(x^2 + 27)^2}$, which has the sign of $x$ (the denominator is a positive square). $f$ decreases on $(-\infty, 0]$, increases on $[0, \infty)$, and since it decreases then increases on the WHOLE line, $f(0) = -\frac{1}{3}$ is the absolute minimum. The form $1 - \frac{36}{x^2 + 27}$ gives the same derivative in one line by the chain rule: $\frac{d}{dx}\left[-36(x^2 + 27)^{-1}\right] = 36(x^2 + 27)^{-2}(2x)$.

c) Quotient rule on $\frac{72x}{(x^2 + 27)^2}$, the derivative of the denominator by the chain rule being $2(x^2 + 27)(2x)$: $f''(x) = \frac{72(x^2 + 27)^2 - 72x \cdot 2(x^2 + 27)(2x)}{(x^2 + 27)^4}$. Both terms of the numerator contain $x^2 + 27$: pull it out, $72(x^2 + 27)\left[(x^2 + 27) - 4x^2\right] = 72(x^2 + 27)(27 - 3x^2)$, and cancel ONE power against the denominator: $f''(x) = \frac{72(27 - 3x^2)}{(x^2 + 27)^3} = \frac{216(9 - x^2)}{(x^2 + 27)^3}$. The expanding route gives the numerator $-216x^4 - 3888x^2 + 52\,488$ over $(x^2 + 27)^4$: correct, but a degree 4 polynomial with large coefficients whose sign must now be found by factoring it back, which is where the marks go. $f''(0) = \frac{216 \times 9}{27^3} = \frac{1944}{19\,683} = \frac{8}{81}$.

d) The denominator $(x^2 + 27)^3$ is positive, so $f''$ has the sign of $9 - x^2 = (3 - x)(3 + x)$: positive on $(-3, 3)$, where $f$ is concave up, negative for $|x| > 3$, where $f$ is concave down. The sign changes at $\pm 3$, which are in the domain: two points of inflection, $(-3, 0)$ and $(3, 0)$. They are also the $x$-intercepts: the graph crosses the axis exactly where it changes its bending. The slopes there: $f'(3) = \frac{216}{36^2} = \frac{216}{1296} = \frac{1}{6}$ and, by symmetry, $f'(-3) = -\frac{1}{6}$. Consistency with the asymptote: an increasing curve that levels off below $y = 1$ must end concave DOWN, which is what $f'' < 0$ for $x > 3$ says.

e) The table from left to right: coming from just below $y = 1$ on the far left, decreasing and concave down; inflection at $(-3, 0)$; concave up down to the minimum $\left(0, -\frac{1}{3}\right)$; up again, inflection at $(3, 0)$; then concave down toward $y = 1$ from below. Range: the minimum $-\frac{1}{3}$ is reached, the value $1$ never is, and every value between is taken by continuity: $\left[-\frac{1}{3}, 1\right)$.

a) $\sqrt{x + 1}$ requires $x + 1 \ge 0$: the domain is $[-1, \infty)$. $f(x) = 0 \iff x = 2$ or $x = -1$, and $f(0) = -2 \cdot 1 = -2$. Since $\sqrt{x + 1} \ge 0$, $f$ has the sign of $x - 2$: negative on $(-1, 2)$, positive after $2$. As $x \to \infty$ both factors tend to $\infty$, so $f \to \infty$. No vertical asymptote ($f$ is continuous on its closed domain), no horizontal one, and no slant one either, because $\frac{f(x)}{x} = \left(1 - \frac{2}{x}\right)\sqrt{x + 1} \to \infty$. There is no symmetry to look for: the domain itself is not symmetric about $0$.

b) Product rule, then chain rule on the root (inner function $x + 1$): $f'(x) = \sqrt{x + 1} + \frac{x - 2}{2\sqrt{x + 1}}$. As a SUM, its sign is unreadable. Over the common denominator: $\sqrt{x + 1} = \frac{2(x + 1)}{2\sqrt{x + 1}}$, so $f'(x) = \frac{2(x + 1) + (x - 2)}{2\sqrt{x + 1}} = \frac{3x}{2\sqrt{x + 1}}$ for $x > -1$. The denominator is positive: $f'$ has the sign of $x$. Critical number $0$ ($f'(0) = 0$). At the endpoint, as $x \to -1^+$ the numerator tends to $-3$ and the denominator to $0^+$: $f'(x) \to -\infty$. The graph leaves $(-1, 0)$ with a VERTICAL tangent, going down.

c) Product rule on $\frac{3}{2}x(x + 1)^{-1/2}$: $f''(x) = \frac{3}{2}\left[(x + 1)^{-1/2} - \frac{1}{2}x(x + 1)^{-3/2}\right]$. The two powers are $-\frac{1}{2}$ and $-\frac{3}{2}$; the SMALLEST is $-\frac{3}{2}$, and $(x + 1)^{-1/2} = (x + 1)^{-3/2}(x + 1)$. So $f''(x) = \frac{3}{2}(x + 1)^{-3/2}\left[(x + 1) - \frac{x}{2}\right] = \frac{3}{2}(x + 1)^{-3/2} \cdot \frac{x + 2}{2} = \frac{3(x + 2)}{4(x + 1)^{3/2}}$. The frequent error is to factor out $(x + 1)^{-1/2}$, the power that looks smaller because $\frac{1}{2} < \frac{3}{2}$: the bracket then keeps a negative power inside, and the sign is still hidden. Now $f''(x) = 0$ gives $x = -2$, but $-2 < -1$ is not in the domain: it is NOT a candidate, and there is no point of inflection. On the whole domain $x + 2 > 1 > 0$, so $f'' > 0$: $f$ is concave up on $(-1, \infty)$.

d) $f''(0) = \frac{3 \cdot 2}{4 \cdot 1} = \frac{3}{2} > 0$ and $f'(0) = 0$: by the Second Derivative Test, $f$ has a local minimum $f(0) = -2$. It is absolute: $f' < 0$ on $(-1, 0)$ and $f' > 0$ on $(0, \infty)$, so $f$ decreases on all of $[-1, 0]$ and increases on all of $[0, \infty)$. The range: from its minimum $-2$ the function increases without bound, and it is continuous, so the range is $[-2, \infty)$. The value at the endpoint, $f(-1) = 0$, is a local maximum in Thomas' sense for an endpoint, but not the largest value.

e) The positive $x$-intercept is $x = 2$: $f'(2) = \frac{3 \cdot 2}{2\sqrt 3} = \frac{3}{\sqrt 3} = \sqrt 3$ (rationalize: $\frac{3}{\sqrt 3} = \frac{3\sqrt 3}{3}$). The sketch: start at $(-1, 0)$ with a vertical tangent, go down to $(0, -2)$ where the tangent is horizontal, come back up through $(2, 0)$ with slope $\sqrt 3$, and rise without bound, bending upward all the way. A drawing with an S shape near $x = -2$ would be drawing a point that does not exist.