Module 13: Linear approximation, extrema, and the Mean Value Theorem

Choose a base point with a known height

Set $f(x)=\sqrt x$ and $a=25$. Then $f(25)=5$ and $f'(25)=1/(2\sqrt{25})=1/10$. Thus $L(x)=5+(x-25)/10$ and $\boxed{\sqrt{25.3}\approx5.03}$. The step is $0.3$, not $25.3$. Concavity makes this an overestimate; only $L(25.3)=5.03$ is exact.

A negative slope does not mean a negative value

The derivative is $-1/x^2$, so $L(x)=1/2-(x-2)/4$. At $2.04$, $L=0.5-0.01=\boxed{0.49}$. Because $f''(x)=2/x^3>0$ for positive $x$, this tangent estimate is below the true reciprocal. As a numerical check, $1/2.04=25/51\approx0.490196$.

Compare all candidates

Continuity ensures attained extrema. Differentiate: $f'=3x^2-6x=3x(x-2)$, with interior critical numbers $0,2$. Evaluate

The absolute maximum is $\boxed1$ at $0,3$; the absolute minimum is $\boxed{-3}$ at $-1,2$. Derivative zeros alone would miss the endpoint ties.

The MVT point need not be unique

The polynomial meets both hypotheses. Its secant slope is $(1-(-1))/(1-(-1))=1$. Solve $3c^2=1$ to obtain $\boxed{c=\pm1/\sqrt3}$. Both lie in $(-1,1)$, so both qualify.

Locate the missing hypothesis

The function is continuous and has equal endpoint values, but is not differentiable at the interior point $0$. On $(-1,0)$ its derivative is $-1$, and on $(0,1)$ it is $1$. It has no input in the open interval with derivative zero. The corner is precisely where the minimum occurs.

Bound an unknown endpoint height

For some $c\in(2,5)$, $(f(5)-7)/3=f'(c)$. Thus $-1\le(f(5)-7)/3\le4$. Multiplying by positive $3$ and adding $7$ gives $\boxed{4\le f(5)\le19}$. The units would be output units; the derivative bound must be multiplied by the interval length.

Two counterexamples to overextended rules

For $f(x)=x^3$, $f'(0)=0$ but the function increases through zero, so there is no local extremum there. For $g(x)=x$ on $(0,1)$, every value has smaller and larger nearby values. Its infimum is $0$ and supremum is $1$, neither attained. The EVT’s closed-interval hypothesis is missing.

A global sensitivity bound from local slopes

When $a=b$, both sides are zero. For $a<b$, sine is continuous and differentiable, so the MVT gives $\sin b-\sin a=\cos c\,(b-a)$ for an intermediate $c$. Taking absolute values yields $|\sin b-\sin a|\le b-a$. Interchanging $a,b$ handles the other order. Therefore a radian input error of magnitude at most $\eta$ changes sine by at most $\eta$; this is an exact bound, not a linear approximation.

A known cube root, then a slope

$f(8)=(8^{1/3})^2=4$. With $\tfrac23-1=\tfrac{2-3}{3}=-\tfrac13$, $f'(x)=\tfrac23x^{-1/3}$, so $f'(8)=\tfrac23\cdot\tfrac12=\tfrac13$. Then $L(x)=4+\tfrac13(x-8)$ and $(8.06)^{2/3}\approx L(8.06)=4+\tfrac{0.06}{3}=\boxed{4.02}$. Next, $f''(x)=-\tfrac29x^{-4/3}<0$ for $x>0$, so the graph is concave down, lies below its tangent, and $4.02$ is too high. A calculator gives $4.019975\ldots$, just below.

Rewrite first, and guard the sign

$f=5x^{-3}$, $f'=-15x^{-4}$, so $f(1)=5$ and $f'(1)=-15$. Then $L(x)=5-15(x-1)$ and $L(1.02)=5-0.3=\boxed{4.70}$. Since $f''=60x^{-5}>0$ for $x>0$, the graph is concave up and lies above its tangent, so $4.70$ is too low. A calculator gives $5/1.061208\approx4.7116$.

Bending decides the error

For sine: $\sin0=0$ and $\sin'(0)=\cos0=1$, so $\sin x\approx x$ and $\sin(0.1)\approx\boxed{0.1}$; the calculator gives $0.0998334$, an error of about $0.00017$. For cosine: $\cos0=1$ and $\cos'(0)=-\sin0=0$, so $\cos x\approx1$ and $\cos(0.1)\approx\boxed{1}$; the calculator gives $0.9950042$, an error of about $0.005$, thirty times larger. The reason is bending: $\sin''(0)=-\sin0=0$, so sine is momentarily straight at $0$ (an inflection point), while $\cos''(0)=-\cos0=-1$, so cosine bends away from its flat tangent at once. Module 30 makes this exact: the error is about $\tfrac12|f''(a)|$ times the square of the step.

Two famous lines and their concavity

$\ln1=0$ and $\tfrac{\dd}{\dd x}\ln x=\tfrac1x$ equals $1$ at $x=1$, so $\ln x\approx x-1$ and $\ln(1.05)\approx\boxed{0.05}$. Since $(\ln x)''=-1/x^2<0$, $\ln$ is concave down and this is too high (true value $0.04879$). Also $e^0=1$ with slope $1$, so $e^x\approx1+x$ and $e^{-0.03}\approx\boxed{0.97}$. Since $(e^x)''=e^x>0$, this is too low (true value $0.970446$).

A radius error becomes three times the relative error

$V=\tfrac43\pi r^3$, so $\dd V=4\pi r^2\,\dd r=4\pi(100)(0.1)=\boxed{40\pi\approx126\text{ cm}^3}$. The volume is $\tfrac{4000\pi}{3}\approx4189$ cm$^3$, so the relative error is $\dfrac{40\pi}{4000\pi/3}=0.03=\boxed{3\%}$, three times the radius’s $1\%$, because the exponent is $3$.

A square root halves relative error

$T=\dfrac{2\pi}{\sqrt g}\,\ell^{1/2}$, so by the exponent rule $\dfrac{\dd T}{T}=\dfrac12\cdot\dfrac{\dd\ell}{\ell}=\dfrac12(0.5\%)=\boxed{0.25\%}$. For $T$ to be good to $0.1\%$, $\ell$ must be good to $\boxed{0.2\%}$, $2$ mm on a $1$ m pendulum.

Three kinds of candidates

$f'(x)=1-x^{-2/3}=1-\dfrac{1}{x^{2/3}}$. It is zero when $x^{2/3}=1$, that is $(x^{1/3})^2=1$, so $x^{1/3}=\pm1$ and $x=\pm1$. It does not exist at $x=0$, which is inside the interval and in $f$’s domain: the formula divides by zero there, and the definition agrees, because $\dfrac{f(h)-f(0)}{h}=1-3h^{-2/3}$ falls without bound as $h\to0$. Candidates: endpoints $-1$ and $8$; $f'=0$ at $1$ (and at $-1$, already listed); $f'$ undefined at $0$.

Absolute maximum $\boxed{2}$ at $x=-1$ and $x=8$; absolute minimum $\boxed{-2}$ at $x=1$.

Two angles from one sine value

$f'(x)=1-2\sin x=0$ when $\sin x=\tfrac12$, which on $[0,\pi]$ happens at $x=\tfrac\pi6$ and $x=\tfrac{5\pi}6$. Compare four heights:

Absolute maximum $\boxed{\tfrac\pi6+\sqrt3}$ at $x=\tfrac\pi6$; absolute minimum $\boxed{\tfrac{5\pi}6-\sqrt3}$ at $x=\tfrac{5\pi}6$. The cosine became $\cos\tfrac\pi6=\tfrac{\sqrt3}{2}$, and the factor $2$ made it $\sqrt3$.

Hypotheses, then one equation

$\sqrt x$ is continuous for $x\ge0$, so on $[1,9]$, and differentiable for $x>0$, so on $(1,9)$. The secant slope is $\dfrac{3-1}{9-1}=\dfrac14$. Solve $\dfrac{1}{2\sqrt c}=\dfrac14$: $\sqrt c=2$, so $\boxed{c=4}$, which lies in $(1,9)$. It is the only one, because $f'(x)=\tfrac{1}{2\sqrt x}$ decreases and so takes each value once.

Squeeze the unknown slope

On $[100,101]$, $\sqrt x$ meets both hypotheses, so $\sqrt{101}-\sqrt{100}=\dfrac{1}{2\sqrt c}(101-100)$ for some $c$ in $(100,101)$. Because $100<c<121$, $10<\sqrt c<11$, so $20<2\sqrt c<22$ and $\dfrac1{22}<\dfrac{1}{2\sqrt c}<\dfrac1{20}$. Therefore $\boxed{\dfrac1{22}<\sqrt{101}-10<\dfrac1{20}}$. (The true value is $0.0498756\ldots$, close to the upper bound because $\sqrt c$ is close to $10$.)

Existence, then uniqueness

$f(x)=x^3+x-1$ is continuous, with $f(0)=-1<0$ and $f(1)=1>0$, so the Intermediate Value Theorem gives a root in $(0,1)$. Suppose there were two roots $a<b$. Then $f(a)=f(b)=0$, and Rolle’s theorem would give $f'(c)=0$ for some $c$ between them. But $f'(x)=3x^2+1\ge1>0$ everywhere, a contradiction. So there is exactly one real root; it is about $0.6823$.

A line through the barometer’s reading

(a) $p'(h)=-\dfrac{p_0}{H}e^{-h/H}$, so $p'(0)=-\dfrac{101{,}325}{8{,}500}\approx\boxed{-11.92\text{ Pa/m}}$ and $L(h)=p_0-11.92h$. The hydrostatic rule gives $-\rho_0g=-1.225(9.81)\approx-12.02$ Pa/m, within $1\%$. That checks the model’s $H$ against an independent number, $\rho_0$: the two agree because $H$ is close to $p_0/(\rho_0g)\approx8.43$ km. It cannot test the one-temperature assumption itself, since the exponential model is built on the same hydrostatic rule.

(b) A drop of $60$ Pa means $h\approx\dfrac{60}{11.92}\approx\boxed{5.0\text{ m}}$.

(c) $\pm\dfrac{3}{11.92}\approx\boxed{\pm0.25\text{ m}}$. A raw barometer altitude wobbles by about a quarter metre, which is why flight controllers filter it and blend it with other sensors (the Kalman filter, module EL22 on the Engineering Lab’s route).

(d) Exact: $p(300)=101{,}325\,e^{-300/8500}\approx\boxed{97{,}811\text{ Pa}}$. The line reaches that pressure at $h=H(1-p/p_0)=8500\bigl(1-e^{-300/8500}\bigr)\approx\boxed{294.8\text{ m}}$, about $5$ m too low. The line lies below the concave-up curve, so it reaches any given pressure too early. Five metres in three hundred, under $2\%$, is fine for a rough altitude, but the error grows like the square of the height (Picture 13.1’s lesson): by $3$ km it is about $470$ m. That is why flight controllers use the full formula.

One slope limit, three consequences

(a) Nesting dolls on $\sigma=(1+e^{-x})^{-1}$: $\sigma'=-(1+e^{-x})^{-2}\cdot(-e^{-x})=\dfrac{e^{-x}}{(1+e^{-x})^2}$. Since $1-\sigma=\dfrac{1+e^{-x}-1}{1+e^{-x}}=\dfrac{e^{-x}}{1+e^{-x}}$, the product $\sigma(1-\sigma)$ is the same fraction, so $\sigma'=\sigma(1-\sigma)$.

(b) $\sigma(0)=\tfrac12$ and $\sigma'(0)=\tfrac12\cdot\tfrac12=\tfrac14$, so $L(x)=\tfrac12+\tfrac x4$ and $\sigma(0.4)\approx\boxed{0.6}$. The true value is $1/(1+e^{-0.4})\approx0.59869$, so the estimate is too high: $\sigma''=\sigma'(1-2\sigma)<0$ when $\sigma>\tfrac12$, that is for $x>0$, so the graph is concave down there.

(c) $\sigma(1-\sigma)=\sigma-\sigma^2=\tfrac14-\left(\sigma-\tfrac12\right)^2\le\tfrac14$, and it is positive because $0<\sigma<1$. If $a=b$, both sides of the inequality are $0$. If $a\ne b$, $\sigma$ is continuous and differentiable everywhere, so the Mean Value Theorem applies on the interval between them: $\sigma(a)-\sigma(b)=\sigma'(c)(a-b)$ for some $c$ in between, and $|\sigma(a)-\sigma(b)|=\sigma'(c)\,|a-b|\le\tfrac14|a-b|$.