Continuity and Differentiability: Mixed Practice — Worked Solutions

Ques­tion 1: Choos­ing $k$ for con­ti­nu­ity#

The prob­lem#

Find $k$ so that $f(x) = \begin{cases} kx^2, & x \le 2 \\ 3, & x > 2 \end{cases}$ is con­tin­u­ous at $2$.

Under­stand­ing the prob­lem#

The func­tion is defined by one rule up to $x = 2$ and another after it. Con­ti­nu­ity at $2$ means the graph has no jump there: the left limit, the right limit and $f(2)$ must all be equal.

The idea#

Use the def­i­n­i­tion: $f$ is con­tin­u­ous at $c$ when $\displaystyle \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$. Write each in terms of $k$ and solve.

Step-by-step solu­tion#

Step 1. Value at $2$. Since $2 \le 2$, the first rule applies.

$$f(2) = k(2)^2 = 4k$$

Step 2. Left limit (val­ues of $x$ just below $2$ use $kx^2$).

$$\displaystyle \lim_{x \to 2^-} kx^2 = 4k$$

Step 3. Right limit (val­ues just above $2$ use the con­stant $3$).

$$\displaystyle \lim_{x \to 2^+} 3 = 3$$

Step 4. For con­ti­nu­ity all three must match.

$$\displaystyle 4k = 3 \quad\Longrightarrow\quad k = \frac{3}{4}$$

Check­ing the answer#

With $\displaystyle k = \frac{3}{4}$: $\displaystyle f(2) = \frac{3}{4} \times 4 = 3$, equal to the right-hand value $3$. ✓

Answer#

$\displaystyle k = \frac{3}{4}$.

Ques­tion 2: Con­ti­nu­ity of $\lvert x \rvert - \lvert x + 1 \rvert$#

The prob­lem#

Dis­cuss the con­ti­nu­ity of $f(x) = \lvert x \rvert - \lvert x + 1 \rvert$.

Under­stand­ing the prob­lem#

"Dis­cuss con­ti­nu­ity" means find where $f$ is con­tin­u­ous and where (if any­where) it is not. The cor­ners of the mod­u­lus graphs at $x = 0$ and $x = -1$ are the places to think about.

The idea#

The mod­u­lus func­tion $\lvert x \rvert$ is con­tin­u­ous every­where, and so is $\lvert x + 1 \rvert$ (a mod­u­lus of a poly­no­mial). The dif­fer­ence of two con­tin­u­ous func­tions is con­tin­u­ous.

Step-by-step solu­tion#

Step 1. $g(x) = \lvert x \rvert$ is con­tin­u­ous at every real num­ber.

Step 2. $h(x) = \lvert x + 1 \rvert$ is the com­pos­ite of the poly­no­mial $x + 1$ and the mod­u­lus func­tion, both con­tin­u­ous, so it is con­tin­u­ous every­where.

Step 3. $f = g - h$ is a dif­fer­ence of con­tin­u­ous func­tions, so it is con­tin­u­ous at every real $x$.

Step 4. A direct check at the cor­ner $x = -1$: for $x < -1$, $f(x) = -x + (x + 1) = 1$; for $-1 \le x < 0$, $f(x) = -x - (x + 1) = -2x - 1$, which gives $1$ at $x = -1$. Both sides agree, so there is no jump. The same hap­pens at $x = 0$, where both pieces give $-1$.

Check­ing the answer#

Con­ti­nu­ity says noth­ing about smooth­ness: the graph has cor­ners at $-1$ and $0$, but no breaks.

Answer#

$f$ is con­tin­u­ous at every real num­ber.

Ques­tion 3: Points of dis­con­ti­nu­ity of a ratio­nal func­tion#

The prob­lem#

Find the points of dis­con­ti­nu­ity of $\displaystyle f(x) = \frac{x^2 + 1}{x^2 - 5x + 6}$.

Under­stand­ing the prob­lem#

A ratio­nal func­tion is con­tin­u­ous wher­ever it is defined. It is unde­fined where the denom­i­na­tor is zero.

The idea#

Fac­torise the denom­i­na­tor and find its zeros.

Step-by-step solu­tion#

Step 1. Fac­torise.

$$x^2 - 5x + 6 = (x - 2)(x - 3)$$

Step 2. The denom­i­na­tor is zero at $x = 2$ and $x = 3$, so $f$ is not defined there.

Step 3. The numer­a­tor $x^2 + 1$ is never zero, so noth­ing can­cels; $f$ blows up at both points. Every­where else $f$ is a quo­tient of con­tin­u­ous poly­no­mi­als with non-zero denom­i­na­tor, hence con­tin­u­ous.

Check­ing the answer#

At $x = 2.01$ the denom­i­na­tor is about $-0.0099$ and $f \approx -509$; at $x = 1.99$, $f \approx +491$. A huge jump, as expected.

Answer#

$f$ is dis­con­tin­u­ous only at $x = 2$ and $x = 3$.

Ques­tion 4: A chain of three func­tions#

The prob­lem#

Dif­fer­en­ti­ate $\sin(\cos(x^2))$.

Under­stand­ing the prob­lem#

This is a com­pos­ite of three func­tions: square, then cosine, then sine. You need $\displaystyle \frac{d}{dx}$ of the whole.

The idea#

Use the chain rule from the out­side in: dif­fer­en­ti­ate the outer func­tion, keep its inside unchanged, then mul­ti­ply by the deriv­a­tive of the inside, and repeat.

Step-by-step solu­tion#

Step 1. Outer func­tion $\sin(\,\cdot\,)$: its deriv­a­tive is $\cos(\,\cdot\,)$.

$$\displaystyle \frac{d}{dx}\sin(\cos x^2) = \cos(\cos x^2) \cdot \frac{d}{dx}\cos(x^2)$$

Step 2. Mid­dle func­tion $\cos(\,\cdot\,)$: its deriv­a­tive is $-\sin(\,\cdot\,)$.

$$\displaystyle \frac{d}{dx}\cos(x^2) = -\sin(x^2) \cdot \frac{d}{dx}(x^2) = -\sin(x^2) \cdot 2x$$

Step 3. Mul­ti­ply.

$$\displaystyle \frac{d}{dx}\sin(\cos x^2) = -2x\sin(x^2)\cos(\cos x^2)$$

Check­ing the answer#

At $x = 0$ the func­tion has a flat point (it depends on $x^2$, an even func­tion), and our deriv­a­tive is $0$ at $x = 0$. ✓

Answer#

$\displaystyle \frac{d}{dx}\sin(\cos x^2) = -2x\sin(x^2)\cos(\cos(x^2))$.

Ques­tion 5: An inverse secant by sub­sti­tu­tion#

The prob­lem#

Dif­fer­en­ti­ate $\displaystyle \sec^{-1}\frac{1}{2x^2 - 1}$ for $\displaystyle 0 < x < \frac{1}{\sqrt{2}}$.

Under­stand­ing the prob­lem#

Dif­fer­en­ti­at­ing directly is messy. The expres­sion $2x^2 - 1$ is the dou­ble-angle form $2\cos^2\theta - 1 = \cos 2\theta$, so a sub­sti­tu­tion will sim­plify it. You must also check the range so that $\sec^{-1}(\sec 2\theta) = 2\theta$ is valid.

The idea#

Put $x = \cos\theta$. Sim­plify the func­tion to a mul­ti­ple of $\theta$, then dif­fer­en­ti­ate using $\theta = \cos^{-1}x$.

Step-by-step solu­tion#

Step 1. Sub­sti­tute $x = \cos\theta$. Since $\displaystyle 0 < x < \frac{1}{\sqrt{2}}$, we have $\displaystyle \frac{\pi}{4} < \theta < \frac{\pi}{2}$.

$$\displaystyle \frac{1}{2x^2 - 1} = \frac{1}{2\cos^2\theta - 1} = \frac{1}{\cos 2\theta} = \sec 2\theta$$

Step 2. Check the range. $\displaystyle \frac{\pi}{2} < 2\theta < \pi$, which lies inside the prin­ci­pal range $[0, \pi]$, $\displaystyle \ne \frac{\pi}{2}$, of $\sec^{-1}$. So

$$y = \sec^{-1}(\sec 2\theta) = 2\theta = 2\cos^{-1}x$$

Step 3. Dif­fer­en­ti­ate, using $\displaystyle \frac{d}{dx}\cos^{-1}x = -\frac{1}{\sqrt{1 - x^2}}$.

$$\displaystyle \frac{dy}{dx} = 2 \times \left(-\frac{1}{\sqrt{1 - x^2}}\right) = -\frac{2}{\sqrt{1 - x^2}}$$

Check­ing the answer#

At $x = 0.5$: $\displaystyle \frac{1}{2(0.25) - 1} = -2$, and a numer­i­cal deriv­a­tive of $\sec^{-1}$ of the expres­sion near $0.5$ gives about $-2.309$, while $\displaystyle -\frac{2}{\sqrt{0.75}} \approx -2.309$. ✓

Answer#

$\displaystyle \frac{dy}{dx} = -\frac{2}{\sqrt{1 - x^2}}$.

Com­mon mis­take to avoid#

Writ­ing $\sec^{-1}(\sec 2\theta) = 2\theta$ with­out check­ing that $2\theta$ lies in the prin­ci­pal range. Here it does; on another inter­val of $x$ the answer would change.

Ques­tion 6: Implicit func­tion $y^x = x^y$#

The prob­lem#

Find $\displaystyle \frac{dy}{dx}$ if $y^x = x^y$.

Under­stand­ing the prob­lem#

Both sides have a vari­able in the base and in the power, and $y$ can­not eas­ily be made the sub­ject. You need $\displaystyle \frac{dy}{dx}$.

The idea#

Take log­a­rithms to bring the pow­ers down, then dif­fer­en­ti­ate both sides implic­itly with respect to $x$, using the prod­uct rule and remem­ber­ing that $y$ is a func­tion of $x$.

Step-by-step solu­tion#

Step 1. Take $\log$ of both sides.

$$x\log y = y\log x$$

Step 2. Dif­fer­en­ti­ate each side with the prod­uct rule.

$$\displaystyle \log y + x \cdot \frac{1}{y}\frac{dy}{dx} = \frac{dy}{dx}\log x + y \cdot \frac{1}{x}$$

Step 3. Col­lect the $\displaystyle \frac{dy}{dx}$ terms on the left.

$$\displaystyle \frac{dy}{dx}\left(\frac{x}{y} - \log x\right) = \frac{y}{x} - \log y$$

Step 4. Divide.

$$\displaystyle \frac{dy}{dx} = \frac{\dfrac{y}{x} - \log y}{\dfrac{x}{y} - \log x}$$

Check­ing the answer#

A sym­bolic check (implicit dif­fer­en­ti­a­tion of $x\log y - y\log x = 0$) gives the same expres­sion. Also, on the line $y = x$ (which sat­is­fies the equa­tion), the for­mula gives $\displaystyle \frac{1 - \log x}{1 - \log x} = 1$, the slope of $y = x$. ✓

Answer#

$\displaystyle \frac{dy}{dx} = \frac{\frac{y}{x} - \log y}{\frac{x}{y} - \log x}$, equiv­a­lently $\displaystyle \frac{y(y - x\log y)}{x(x - y\log x)}$.

Ques­tion 7: A para­met­ric deriv­a­tive#

The prob­lem#

Find $\displaystyle \frac{dy}{dx}$ if $x = \cos\theta + \theta\sin\theta$, $y = \sin\theta - \theta\cos\theta$.

Under­stand­ing the prob­lem#

$x$ and $y$ are each given in terms of a para­me­ter $\theta$. You need the slope $\displaystyle \frac{dy}{dx}$.

The idea#

Use the para­met­ric rule $\displaystyle \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$. Each deriv­a­tive needs the prod­uct rule.

Step-by-step solu­tion#

Step 1. Dif­fer­en­ti­ate $x$.

$$\displaystyle \frac{dx}{d\theta} = -\sin\theta + (\sin\theta + \theta\cos\theta) = \theta\cos\theta$$

Step 2. Dif­fer­en­ti­ate $y$.

$$\displaystyle \frac{dy}{d\theta} = \cos\theta - (\cos\theta - \theta\sin\theta) = \theta\sin\theta$$

Step 3. Divide.

$$\displaystyle \frac{dy}{dx} = \frac{\theta\sin\theta}{\theta\cos\theta} = \tan\theta$$

Check­ing the answer#

The $\theta$ can­cels only when $\theta \ne 0$; at $\theta = 0$ both deriv­a­tives are $0$ and the rule does not apply there. Else­where the result $\tan\theta$ is valid.

Answer#

$\displaystyle \frac{dy}{dx} = \tan\theta$.

Ques­tion 8: A vari­able power $(\log x)^{\cos x}$#

The prob­lem#

Dif­fer­en­ti­ate $(\log x)^{\cos x}$.

Under­stand­ing the prob­lem#

The base and the expo­nent both depend on $x$, so nei­ther the power rule nor the expo­nen­tial rule applies directly. (We need $\log x > 0$, that is $x > 1$.)

The idea#

Log­a­rith­mic dif­fer­en­ti­a­tion: put $y = (\log x)^{\cos x}$, take $\log$, dif­fer­en­ti­ate, then mul­ti­ply back by $y$.

Step-by-step solu­tion#

Step 1. Take log­a­rithms.

$$\log y = \cos x \cdot \log(\log x)$$

Step 2. Dif­fer­en­ti­ate both sides. On the right use the prod­uct rule and the chain rule for $\log(\log x)$.

$$\displaystyle \frac{1}{y}\frac{dy}{dx} = -\sin x\,\log(\log x) + \cos x \cdot \frac{1}{\log x} \cdot \frac{1}{x}$$

Step 3. Mul­ti­ply by $y$.

$$\displaystyle \frac{dy}{dx} = (\log x)^{\cos x}\left(\frac{\cos x}{x\log x} - \sin x\,\log(\log x)\right)$$

Check­ing the answer#

A sym­bolic check with sympy gives a dif­fer­ence of $0$ from this expres­sion.

Answer#

$\displaystyle \frac{d}{dx}(\log x)^{\cos x} = (\log x)^{\cos x}\left(\frac{\cos x}{x\log x} - \sin x\,\log(\log x)\right)$.

Ques­tion 9: A sec­ond-order rela­tion#

The prob­lem#

If $y = 3\cos(\log x) + 4\sin(\log x)$, show that $x^2 y'' + x y' + y = 0$.

Under­stand­ing the prob­lem#

You must find $y'$ and $y''$ and show that the given com­bi­na­tion is zero. Expand­ing $y''$ directly is messy; there is a neater route.

The idea#

Find $y'$, mul­ti­ply by $x$ to remove the frac­tion, and then dif­fer­en­ti­ate $xy'$ using the prod­uct rule. That pro­duces $xy'' + y'$ directly.

Step-by-step solu­tion#

Step 1. Dif­fer­en­ti­ate with the chain rule ($\displaystyle \frac{d}{dx}\log x = \frac{1}{x}$).

$$\displaystyle y' = -3\sin(\log x) \cdot \frac{1}{x} + 4\cos(\log x) \cdot \frac{1}{x}$$

Step 2. Mul­ti­ply by $x$.

$$xy' = -3\sin(\log x) + 4\cos(\log x)$$

Step 3. Dif­fer­en­ti­ate both sides. The left side needs the prod­uct rule.

$$\displaystyle xy'' + y' = -3\cos(\log x) \cdot \frac{1}{x} - 4\sin(\log x) \cdot \frac{1}{x}$$

Step 4. The right side is $\displaystyle -\frac{1}{x}\big(3\cos(\log x) + 4\sin(\log x)\big) = -\frac{y}{x}$.

$$\displaystyle xy'' + y' = -\frac{y}{x}$$

Step 5. Mul­ti­ply by $x$ and rearrange.

$$x^2y'' + xy' + y = 0$$

Check­ing the answer#

Sub­sti­tut­ing $y$ into $x^2y'' + xy' + y$ sym­bol­i­cally sim­pli­fies to $0$.

Answer#

$x^2 y'' + x y' + y = 0$, as required.

Ques­tion 10: A trigono­met­ric sub­sti­tu­tion#

The prob­lem#

If $\sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y)$, show that $\displaystyle \frac{dy}{dx} = \sqrt{\frac{1 - y^2}{1 - x^2}}$. (Put $x = \sin A$, $y = \sin B$.)

Under­stand­ing the prob­lem#

Dif­fer­en­ti­at­ing the equa­tion as it stands is pos­si­ble but heavy. The hint sug­gests a sub­sti­tu­tion that turns the square roots into cosines. The goal is to show that $A - B$ is a con­stant.

The idea#

With $x = \sin A$, $y = \sin B$ we get $\sqrt{1 - x^2} = \cos A$ and $\sqrt{1 - y^2} = \cos B$ (tak­ing $A, B$ in $\displaystyle \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$). Use the sum-to-prod­uct for­mu­las to sim­plify, then dif­fer­en­ti­ate $\sin^{-1}x - \sin^{-1}y = \text{constant}$.

Step-by-step solu­tion#

Step 1. Sub­sti­tute.

$$\cos A + \cos B = a(\sin A - \sin B)$$

Step 2. Use sum-to-prod­uct for­mu­las.

$$\displaystyle 2\cos\frac{A + B}{2}\cos\frac{A - B}{2} = a \cdot 2\cos\frac{A + B}{2}\sin\frac{A - B}{2}$$

Step 3. Can­cel $\displaystyle 2\cos\frac{A + B}{2}$ (non-zero).

$$\displaystyle \cos\frac{A - B}{2} = a\sin\frac{A - B}{2} \quad\Longrightarrow\quad \cot\frac{A - B}{2} = a$$

Step 4. So $\displaystyle \frac{A - B}{2} = \cot^{-1}a$, a con­stant. Hence

$$\sin^{-1}x - \sin^{-1}y = 2\cot^{-1}a = \text{constant}$$

Step 5. Dif­fer­en­ti­ate with respect to $x$.

$$\displaystyle \frac{1}{\sqrt{1 - x^2}} - \frac{1}{\sqrt{1 - y^2}}\frac{dy}{dx} = 0$$

Step 6. Solve for $\displaystyle \frac{dy}{dx}$.

$$\displaystyle \frac{dy}{dx} = \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}} = \sqrt{\frac{1 - y^2}{1 - x^2}}$$

Check­ing the answer#

Numer­i­cally, with $a = 2$ and $x = 0.5$, solv­ing the equa­tion for $y$ and com­put­ing the implicit slope gives $1.0619$, and $\displaystyle \sqrt{\frac{1 - y^2}{1 - x^2}}$ gives the same $1.0619$. ✓

Answer#

$\displaystyle \frac{dy}{dx} = \sqrt{\frac{1 - y^2}{1 - x^2}}$, as required.

Ques­tion 11: A sum of two vari­able pow­ers#

The prob­lem#

Dif­fer­en­ti­ate $x^{x^2 - 3} + (x - 3)^{x^2}$ for $x > 3$.

Under­stand­ing the prob­lem#

Each term has a vari­able base and a vari­able expo­nent. You can­not take the log­a­rithm of a sum use­fully, so treat the two terms sep­a­rately.

The idea#

Write $u = x^{x^2 - 3}$ and $v = (x - 3)^{x^2}$. Then $\displaystyle \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$, and find each by log­a­rith­mic dif­fer­en­ti­a­tion. The con­di­tion $x > 3$ makes $\log(x - 3)$ defined.

Step-by-step solu­tion#

Step 1. For $u$: take log­a­rithms and dif­fer­en­ti­ate.

$$\displaystyle \begin{aligned} \log u &= (x^2 - 3)\log x \\ \frac{1}{u}\frac{du}{dx} &= 2x\log x + (x^2 - 3)\cdot\frac{1}{x} \\ \frac{du}{dx} &= x^{x^2 - 3}\left(\frac{x^2 - 3}{x} + 2x\log x\right) \end{aligned}$$

Step 2. For $v$: the same method.

$$\displaystyle \begin{aligned} \log v &= x^2\log(x - 3) \\ \frac{1}{v}\frac{dv}{dx} &= 2x\log(x - 3) + x^2 \cdot \frac{1}{x - 3} \\ \frac{dv}{dx} &= (x - 3)^{x^2}\left(\frac{x^2}{x - 3} + 2x\log(x - 3)\right) \end{aligned}$$

Step 3. Add.

$$\displaystyle \frac{dy}{dx} = x^{x^2 - 3}\left(\frac{x^2 - 3}{x} + 2x\log x\right) + (x - 3)^{x^2}\left(\frac{x^2}{x - 3} + 2x\log(x - 3)\right)$$

Check­ing the answer#

At $x = 4.3$ the for­mula matches the deriv­a­tive com­puted by sympy (dif­fer­ence $0$).

Answer#

$\displaystyle \frac{dy}{dx} = x^{x^2 - 3}\left(\frac{x^2 - 3}{x} + 2x\log x\right) + (x - 3)^{x^2}\left(\frac{x^2}{x - 3} + 2x\log(x - 3)\right)$.

Com­mon mis­take to avoid#

Tak­ing $\log$ of the whole sum and writ­ing $\log(u + v) = \log u + \log v$. That is false; split the sum first.

Ques­tion 12: Con­tin­u­ous but not dif­fer­en­tiable#

The prob­lem#

Show that $f(x) = \lvert x - 1 \rvert + \lvert x - 2 \rvert$ is con­tin­u­ous every­where but not dif­fer­en­tiable at $1$ and $2$.

Under­stand­ing the prob­lem#

Two things must be shown: (i) no breaks any­where; (ii) at $x = 1$ and $x = 2$ the left-hand and right-hand deriv­a­tives dif­fer (the graph has cor­ners).

The idea#

Con­ti­nu­ity fol­lows because a sum of con­tin­u­ous func­tions is con­tin­u­ous. For dif­fer­en­tia­bil­ity, remove the mod­u­lus signs on each inter­val to get a straight-line for­mula, then com­pare slopes on either side of $1$ and of $2$.

Step-by-step solu­tion#

Step 1. Con­ti­nu­ity. $\lvert x - 1 \rvert$ and $\lvert x - 2 \rvert$ are each con­tin­u­ous every­where, so their sum $f$ is con­tin­u­ous every­where.

Step 2. Write $f$ with­out mod­u­lus signs.

$$f(x) = \begin{cases} (1 - x) + (2 - x) = 3 - 2x, & x < 1 \\ (x - 1) + (2 - x) = 1, & 1 \le x \le 2 \\ (x - 1) + (x - 2) = 2x - 3, & x > 2 \end{cases}$$

Step 3. At $x = 1$: the left-hand deriv­a­tive is the slope of $3 - 2x$, which is $-2$; the right-hand deriv­a­tive is the slope of the con­stant $1$, which is $0$.

$$Lf'(1) = -2 \ne 0 = Rf'(1)$$

So $f$ is not dif­fer­en­tiable at $1$.

Step 4. At $x = 2$: left-hand deriv­a­tive $0$ (con­stant piece), right-hand deriv­a­tive $2$ (slope of $2x - 3$).

$$Lf'(2) = 0 \ne 2 = Rf'(2)$$

So $f$ is not dif­fer­en­tiable at $2$.

Check­ing the answer#

The pieces meet: at $x = 1$, $3 - 2 = 1$; at $x = 2$, $4 - 3 = 1$. So the graph is joined (con­tin­u­ous) but bends sharply at both points.

Answer#

$f$ is con­tin­u­ous every­where; at $x = 1$ the one-sided deriv­a­tives are $-2$ and $0$, and at $x = 2$ they are $0$ and $2$, so $f$ is not dif­fer­en­tiable at $1$ or $2$.