Tag Archives: Feliciano and Uy

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 2

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PROBLEM:

Evaluate lim⁡x→3(4x+2x+4)\displaystyle \lim\limits_{x\to 3}\left(\frac{4x+2}{x+4}\right).

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SOLUTION:

lim⁡x→3(4x+2x+4)=lim⁡x→3(4x+2)lim⁡x→3(x+4)=lim⁡x→3(4x)+lim⁡x→3(2)lim⁡x→3(x)+lim⁡x→3(4)=4⋅lim⁡x→3(x)+23+4=4⋅3+23+4=12+27=147=2  (Answer)\begin{align*} \lim_{x\to 3}\left(\frac{4x+2}{x+4}\right)& =\frac{\lim\limits_{x\to 3}\left(4x+2\right)}{\lim\limits_{x\to 3}\left(x+4\right)}\\ & =\frac{\lim\limits_{x\to 3}\left(4x\right)+\lim\limits_{x\to 3}\left(2\right)}{\lim\limits_{x\to 3}\left(x\right)+\lim\limits_{x\to 3}\left(4\right)}\\ & =\frac{4\cdot \lim\limits_{x\to 3}\left(x\right)+2}{3+4}\\ & =\frac{4\cdot 3+2}{3+4}\\ & =\frac{12+2}{7}\\ & =\frac{14}{7}\\ & =2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 1

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PROBLEM:

Evaluate lim⁡x→2(x2−4x+3)\displaystyle \lim _{x\to 2}\left(x^2-4x+3\right).

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SOLUTION:

lim⁡x→2(x2−4x+3)=lim⁡x→2(x2)−lim⁡x→2(4x)+lim⁡x→2(3)=[lim⁡x→2(x)]2−4lim⁡x→2(x)+3=(2)2−4(2)+3=−1  (Answer)\begin{align*} \lim_{x\to 2}\left(x^2-4x+3\right)& = \lim_{x\to 2}\left(x^2\right)-\lim_{x\to 2}\left(4x\right)+\lim_{x\to 2}\left(3\right)\\ & =\left[\lim_{x\to 2}\left(x\right)\right]^2-4\lim_{x\to 2}\left(x\right)+3\\ & =\left(2\right)^2-4\left(2\right)+3\\ & =-1 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1, Problem 10

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PROBLEM:

If  f(x)=4x+3\displaystyle f\left(x\right)=\frac{4}{x+3} and  g(x)=x2−3\displaystyle \:g\left(x\right)=x^2-3 , find f[g(x)]\displaystyle f\left[g\left(x\right)\right] and g[f(x)]\displaystyle g\left[f\left(x\right)\right].

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SOLUTION:

Part A

f[g(x)]=4(x2−3)+3=4x2  (Answer)\begin{align*} f\left[g\left(x\right)\right] & =\frac{4}{\left(x^2-3\right)+3}\\ & =\frac{4}{x^2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}

Part B

g[f(x)]=(4x+3)2−3=16(x+3)2−3=16−3(x+3)2(x+3)2=16−3(x2+6x+9)(x+3)2=16−3x2−18x−27(x+3)2=−3x2−18x−11(x+3)2  (Answer)\begin{align*} g\left[f\left(x\right)\right] & =\left(\frac{4}{x+3}\right)^2-3\\ & =\frac{16}{\left(x+3\right)^2}-3\\ & =\frac{16-3\left(x+3\right)^2}{\left(x+3\right)^2}\\ & =\frac{16-3\left(x^2+6x+9\right)}{\left(x+3\right)^2}\\ & =\frac{16-3x^2-18x-27}{\left(x+3\right)^2}\\ & =\frac{-3x^2-18x-11}{\left(x+3\right)^2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1, Problem 9

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PROBLEM:

If f(x)=3x2−4x+1\displaystyle f\left(x\right)=3x^2-4x+1, find f(h+3)−f(3)h, h≠0\displaystyle \frac{f\left(h+3\right)-f\left(3\right)}{h},\:h\ne 0.

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SOLUTION:

f(h+3)−f(3)h=[3(h+3)2−4(h+3)+1]−[3(3)2−4(3)+1]h=3(h2+6h+9)−4h−12+1−16h=3h2+18h+27−4h−12+1−16h=3h2+14hh=h(3h+14)h=3h+14  (Answer)\begin{align*} \frac{f\left(h+3\right)-f\left(3\right)}{h} & =\frac{\left[3\left(h+3\right)^2-4\left(h+3\right)+1\right]-\left[3\left(3\right)^2-4\left(3\right)+1\right]}{h} \\ & =\frac{3\left(h^2+6h+9\right)-4h-12+1-16}{h}\\ & =\frac{3h^2+18h+27-4h-12+1-16}{h}\\ & =\frac{3h^2+14h}{h}\\ & =\frac{h\left(3h+14\right)}{h}\\ & =3h+14 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1, Problem 8

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PROBLEM:

If f(x)=x2+1\displaystyle f\left(x\right)=x^2+1, find f(x+h)−f(x)h, h≠0\displaystyle \frac{f\left(x+h\right)-f\left(x\right)}{h},\:h\ne 0.

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SOLUTION:

f(x+h)−f(x)h=[(x+h)2+1]−(x2+1) h=x2+2xh+h2+1−x2−1h=2xh+h2h=h(2x+h)h=2x+h  (Answer)\begin{align*} \displaystyle \frac{f\left(x+h\right)-f\left(x\right)}{h} & =\frac{\left[\left(x+h\right)^2+1\right]-\left(x^2+1\right)\:}{h}\\ \\ & =\frac{x^2+2xh+h^2+1-x^2-1}{h}\\ \\ & =\frac{2xh+h^2}{h}\\ \\ & =\frac{h\left(2x+h\right)}{h}\\ \\ & =2x+h \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1, Problem 7

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PROBLEM:

A right circular cylinder, a radius of base xx, height yy, is inscribed in a right circular cone, radius of base rr and a height hh. Express yy as a function of xx (rr and hh are constants).

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SOLUTION:

Refer to the figure below for an elevation view.

Schematic Diagram of a right circular cylinder inscribed in a right circular cone.
Diagram of a right circular cylinder with a base radius of r and height y inscribed in a right circular cone with base radius r and height h.

By ratio and proportion of two similar triangles, we have

yr−x=hry=h(r−x) r  (Answer)\begin{align*} \frac{y}{r-x} & = \frac{h}{r} \\ y & =\frac{h\left(r-x\right)\:}{r} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1 Problem 6

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PROBLEM:

The stiffness of a beam of rectangular cross-section is proportional to the breadth and the cube of the depth. If the breadth is 20 cm, express the stiffness as a function of the depth.

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SOLUTION:

Let SS=stiffness, bb=breadth, and dd=depth

S=bd3S=20d3  (Answer)\begin{align*} S & =bd^3 \\ S & = 20 d^3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1 Problem 5

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PROBLEM:

Express the area AA of an equilateral triangle as a function of its side xx.

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SOLUTION:

From the formula of the area of a triangle, A=12absin⁡(θ)\displaystyle A=\frac{1}{2} \text{a}\text{b} \sin\left(\theta \right). Also, we know that the interior angle of an equilateral triangle is 60 degrees, and sin⁡ 60∘=32\displaystyle \sin\:60^{\circ} =\frac{\sqrt{3}}{2}.

A=12absin⁡(θ)A=12⋅x⋅x⋅sin⁡ 60∘A=12⋅x2⋅32A=34x2  (Answer)\begin{align*} A & =\frac{1}{2} \text{a}\text{b} \sin\left(\theta \right) \\ A & =\frac{1}{2} \cdot x\cdot x\cdot \sin\:60^{\circ} \\ A & =\frac{1}{2}\cdot x^2\cdot \frac{\sqrt{3}}{2} \\ A & =\frac{\sqrt{3}}{4}x^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1 Problem 4

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PROBLEM:

Express the distance DD traveled in tt hr by a car whose speed is 60 km/hr.

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SOLUTION:

Distance=Rate×TimeD=(60 km/hr)⋅t hrD=60t km  (Answer)\begin{align*} \text{Distance} & = \text{Rate} \times \text{Time} \\ D & =\left(60\:\text{km/hr} \right)\cdot t \ \text{hr} \\ D & =60t \ \text{km} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1 Problem 3

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PROBLEM:

If y=tan⁡(x+π) \displaystyle y= \tan\left(x+\pi \right), find xx as a function of yy.

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SOLUTION:

y=tan⁡(x+π)x+π=tan⁡−1yx=tan⁡−1y−π  (Answer)\begin{align*} y & = \tan\left(x+\pi \right) \\ x+\pi & = \tan^{-1}y \\ x & = \tan^{-1}y-\pi \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
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