Tag Archives: Differential and Integral Calculus by Feliciano and Uy Solution Manual

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 6

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PROBLEM:

Evaluate lim⁡x→2(4x−3)(x2+5)\displaystyle \lim_{x\to 2}\left(4x-3\right)\left(x^2+5\right).

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SOLUTION:

Plug the value x=2.

lim⁡x→2(4x−3)(x2+5)=[(4⋅2)−3][(2)2+5]=[8−3][4+5]=(5)(9)=45  (Answer)\begin{align*} \lim\limits_{x\to 2}\left(4x-3\right)\left(x^2+5\right) & =\left[\left(4\cdot 2\right)-3\right]\left[\left(2\right)^2+5\right]\\ & =\left[8-3\right]\left[4+5\right]\\ & =\left(5\right)\left(9\right)\\ & =45 \ \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 5

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PROBLEM:

Evaluate lim⁡x→8(2x+x3−4)\displaystyle \lim\limits_{x\to 8}\left(2x+\sqrt[3]{x}-4\right).

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SOLUTION:

Plug in the value x=8.

lim⁡x→8(2x+x3−4)=[2(8)+83−4]=[16+2−4]=14  (Answer)\begin{align*} \lim\limits_{x\to 8}\left(2x+\sqrt[3]{x}-4\right) & = \left[2\left(8\right)+\sqrt[3]{8}-4\right]\\ & =\left[16+2-4\right]\\ & =14 \ \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 4

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PROBLEM:

Evaluate lim⁡x→π3(sin⁡ 2xsin⁡ x)\displaystyle \lim\limits _{x\to \frac{\pi }{3}}\left(\frac{\sin\:2x}{\sin\:x}\right).

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SOLUTION:

Plug in the value x=π3\displaystyle x=\frac{\pi }{3}.

lim⁡x→π3(sin⁡ 2xsin⁡ x)=sin⁡(2⋅π3)sin⁡ (π3)=3232=1  (Answer)\begin{align*} \lim\limits_{x\to \frac{\pi }{3}}\left(\frac{\sin\:2x}{\sin\:x}\right) & =\frac{\sin\left(2\cdot \frac{\pi }{3}\right)}{\sin\:\left(\frac{\pi }{3}\right)} \\ & =\frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}}\\ & =1 \ \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 3

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PROBLEM:

Evaluate lim⁡x→π4(tan⁡ x+sin⁡ x)\displaystyle \lim\limits_{x\to \frac{\pi }{4}}\left(\tan\:x+\sin\:x\right).

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SOLUTION:

lim⁡x→π4(tan⁡ x+sin⁡ x)=lim⁡x→π4(tan⁡ x)+lim⁡x→π4(sin⁡ x)=tan⁡ π4+sin⁡ π4=1+22=2+22  (Answer)\begin{align*} \lim\limits_{x\to \frac{\pi }{4}}\left(\tan\:x+\sin\:x\right) & =\lim\limits_{x\to \frac{\pi }{4}}\left(\tan\:x\right)+\lim\limits_{x\to \frac{\pi }{4}}\left(\sin\:x\right)\\ & =\tan\:\frac{\pi }{4}+\sin\:\frac{\pi }{4}\\ & =1+\frac{\sqrt{2}}{2}\\ & =\frac{2+\sqrt{2}}{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 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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