Applications of Derivatives: practice solutions, College – download the PDF

1 Where does it rise? ★★★

\(f^{\prime}(x)=2x-6=0\) gives \(x=3\). For \(x<3\), \(f^{\prime}<0\): decreasing on \((-\infty,3)\). For \(x>3\), \(f^{\prime}>0\): increasing on \((3,\infty)\).

The sign changes from \(-\) to \(+\), so \(f(3)=9-18+5=-4\) is a local (and absolute) minimum.

2 True or false? ★★★

(a) True: a positive derivative means positive tangent slopes, so \(f\) increases on the whole interval.

(b) False: take \(f(x)=(x-2)^3\). Then \(f^{\prime}(2)=0\), but \(f\) is increasing on both sides of \(2\), so there is no extremum.

3 A cubic hill and valley ★★★

\(g^{\prime}(x)=3x^2-12=3(x-2)(x+2)\), so the critical points are \(x=-2\) and \(x=2\).

\(g^{\prime}>0\) for \(x<-2\), \(g^{\prime}<0\) on \((-2,2)\), \(g^{\prime}>0\) for \(x>2\). So \(g(-2)=-8+24=16\) is a local maximum and \(g(2)=8-24=-16\) is a local minimum.

4 Concavity check ★★★

\(f^{\prime}(x)=3x^2-18x\) and \(f^{\prime\prime}(x)=6x-18\), which is zero at \(x=3\).

\(f^{\prime\prime}<0\) for \(x<3\): concave down; \(f^{\prime\prime}>0\) for \(x>3\): concave up. The concavity changes, so the inflection point is \((3,f(3))=(3,27-81+2)=(3,-52)\).

5 Estimating a square ★★★

\(f(10)=100\) and \(f^{\prime}(10)=20\), so \(L(x)=100+20(x-10)\).

\(L(10.2)=100+20(0.2)=104\). The exact value is \(10.2^2=104.04\), so the error is \(0.04\).

6 A hard-looking limit ★★★

At \(x=3\) the form is \(\dfrac00\). So the limit equals \(\displaystyle\lim_{x\to3}\dfrac{2x}{1}=6\).

Check: \(\dfrac{x^2-9}{x-3}=x+3\) for \(x\ne3\), which tends to \(6\).

7 A growing square ★★★

\(A=s^2\), so \(\dfrac{dA}{dt}=2s\dfrac{ds}{dt}\).

When \(s=10\): \(\dfrac{dA}{dt}=2(10)(3)=60\) cm\(^2\)/s.

8 The open box ★★★

Base side \(12-2x\), height \(x\): \(V(x)=x(12-2x)^2=4x^3-48x^2+144x\) with \(0

\(V^{\prime}(x)=12x^2-96x+144=12(x-2)(x-6)\). Only \(x=2\) lies in \((0,6)\). \(V\) increases before \(2\) and decreases after, and \(V\) is \(0\) at both ends.

The maximum volume is \(V(2)=2\cdot8^2=128\text{ in}^3\) (about \(2{,}098\text{ cm}^3\)), for \(x=2\) in.

9 Bending quartic ★★★

\(f^{\prime}(x)=4x^3-12x\), \(f^{\prime\prime}(x)=12x^2-12=12(x-1)(x+1)\).

\(f^{\prime\prime}>0\) for \(x<-1\) and \(x>1\) (concave up); \(f^{\prime\prime}<0\) on \((-1,1)\) (concave down). Inflection points: \((-1,-5)\) and \((1,-5)\), since \(f(\pm1)=1-6=-5\).

10 Mean Value Theorem for a cube ★★★

\(f\) is a polynomial, so it is continuous on \([0,3]\) and differentiable on \((0,3)\).

The average rate is \(\dfrac{27-0}{3-0}=9\). Solve \(3c^2=9\): \(c=\pm\sqrt3\). Only \(c=\sqrt3\approx1.732\) lies in \((0,3)\).

11 The turnpike driver ★★★

Position \(s(t)\) is continuous and differentiable on the 2-hour interval. The average speed is \(\dfrac{170-20}{2}=75\) mph.

By the Mean Value Theorem, there is a time \(c\) between 1:00 pm and 3:00 pm with \(s^{\prime}(c)=75\): the speedometer read exactly 75 mph (about 120.7 km/h) at that moment.

12 Twice L’Hôpital ★★★

At \(x=0\) the form is \(\dfrac00\). First application: \(\dfrac{2e^{2x}-2}{2x}\), still \(\dfrac00\). Second application: \(\dfrac{4e^{2x}}{2}\to\dfrac42=2\).

The limit is \(2\).

13 The sliding ladder ★★★

\(x^2+y^2=100\), so \(x\dfrac{dx}{dt}+y\dfrac{dy}{dt}=0\). With \(x=8\), \(y=6\): \(\dfrac{dy}{dt}=-\dfrac{8}{6}(1.5)=-2\) ft/s.

The top moves down at 2 ft/s; the negative sign means \(y\) is decreasing.

14 A cube root estimate ★★★

\(f(27)=3\) and \(f^{\prime}(x)=\dfrac13x^{-2/3}\), so \(f^{\prime}(27)=\dfrac{1}{3\cdot9}=\dfrac1{27}\).

\(L(28)=3+\dfrac1{27}\approx3.0370\). The calculator value is \(3.0366\), so the estimate is excellent.

15 The quartic that fooled the test ★★★

\(f^{\prime}(x)=4x^3-12x^2=4x^2(x-3)\): critical points \(x=0\) and \(x=3\). \(f^{\prime}<0\) for \(x<3\) except at \(0\) (where it is \(0\)), and \(f^{\prime}>0\) for \(x>3\). So there is no extremum at \(0\), and \(f(3)=81-108=-27\) is a local minimum.

\(f^{\prime\prime}(x)=12x^2-24x=12x(x-2)\): positive for \(x<0\), negative on \((0,2)\), positive for \(x>2\). Inflection points: \((0,0)\) and \((2,-16)\).

At \(x=0\) the second derivative test fails because \(f^{\prime\prime}(0)=0\); the first derivative test shows the sign of \(f^{\prime}\) does not change.

16 Closest point on a curve ★★★

Minimize the squared distance \(D(x)=(x-4)^2+(\sqrt x)^2=(x-4)^2+x\) for \(x\ge0\).

\(D^{\prime}(x)=2(x-4)+1=2x-7=0\) gives \(x=3.5\). \(D\) decreases before and increases after, and \(D(0)=16\) is larger. \(D(3.5)=0.25+3.5=3.75\).

The closest point is \((3.5,\sqrt{3.5})\approx(3.5,1.871)\), at distance \(\sqrt{3.75}\approx1.94\).

17 The cheapest can ★★★

\(\pi r^2h=500\) gives \(h=\dfrac{500}{\pi r^2}\). Then \(S(r)=2\pi r^2+2\pi rh=2\pi r^2+\dfrac{1000}{r}\).

\(S^{\prime}(r)=4\pi r-\dfrac{1000}{r^2}=0\) gives \(r^3=\dfrac{250}{\pi}\), so \(r=\sqrt[3]{250/\pi}\approx4.30\) cm. Since \(S^{\prime\prime}(r)=4\pi+\dfrac{2000}{r^3}>0\), this is a minimum.

Then \(h=\dfrac{500}{\pi r^2}\approx8.60\) cm, which is \(2r\): the best can is as tall as it is wide.

18 Filling a cone ★★★

By similar triangles \(r=\dfrac h2\), so \(V=\dfrac13\pi r^2h=\dfrac{\pi h^3}{12}\).

\(\dfrac{dV}{dt}=\dfrac{\pi h^2}{4}\dfrac{dh}{dt}\). At \(h=4\): \(4=\dfrac{16\pi}{4}\dfrac{dh}{dt}=4\pi\dfrac{dh}{dt}\), so \(\dfrac{dh}{dt}=\dfrac1\pi\approx0.318\) ft/min (about 9.7 cm/min).

19 Two indeterminate forms ★★★

(a) \(x^2e^{-x}=\dfrac{x^2}{e^x}\), form \(\dfrac\infty\infty\). Apply twice: \(\dfrac{2x}{e^x}\), then \(\dfrac{2}{e^x}\to0\). The limit is \(0\).

(b) \(x\ln x=\dfrac{\ln x}{1/x}\), form \(\dfrac{-\infty}{\infty}\). Then \(\dfrac{1/x}{-1/x^2}=-x\to0\). The limit is \(0\).

20 An inequality from the MVT ★★★

\(f\) is continuous on \([0,x]\) and differentiable on \((0,x)\). The MVT gives \(c\in(0,x)\) with \(\dfrac{\ln(1+x)-\ln1}{x}=f^{\prime}(c)=\dfrac1{1+c}\).

Since \(c>0\), \(\dfrac1{1+c}<1\), so \(\ln(1+x)

Check: \(\ln1.5\approx0.405<0.5\).

21 Find the cubic ★★★

\(f^{\prime}(x)=3x^2+2ax+b\) must vanish at \(-1\) and \(3\), so \(f^{\prime}(x)=3(x+1)(x-3)=3x^2-6x-9\). Hence \(a=-3\) and \(b=-9\), and \(f(x)=x^3-3x^2-9x\).

\(f(-1)=-1-3+9=5\) (local maximum) and \(f(3)=27-27-27=-27\) (local minimum).

\(f^{\prime\prime}(x)=6x-6\) changes sign at \(x=1\); \(f(1)=1-3-9=-11\), so the inflection point is \((1,-11)\).