Tag Archives: Differential Calculus by Feliciano and Uy

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

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PROBLEM:

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

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SOLUTION:

Plug the value x=2.

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

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SOLUTION:

Plug in the value x=8.

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

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SOLUTION:

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

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

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SOLUTION:

\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 1

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PROBLEM:

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

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SOLUTION:

\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 8

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PROBLEM:

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

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SOLUTION:

\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 2

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PROBLEM:

If \displaystyle y=\frac{x^2+3}{x}, find x as a function of y.

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SOLUTION:

\begin{align*}
y & = \frac{x^2+3}{x} \\
xy & =x^2+3 \\
x^2-xy+3&=0 
\end{align*}

Solve for x using the quadratic formula. We have a=1,\:b=-y,\:\text{and}\:c=3

\begin{align*}
x & =\frac{-b\pm \sqrt{b^2-4ac}\:}{2a} \\
x & =\frac{ -\left(-y\right)\pm \sqrt{\left(-y\right)^2-4\left(1\right)\left(3\right)}}{2\left(1\right)} \\
x & =\frac{y\pm \sqrt{y^2-12}}{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 1

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PROBLEM:

If \displaystyle f\left(x\right)=x^2-4x, find

a) \displaystyle f\left(-5\right)

b) \displaystyle f\left(y^2+1\right)

c) \displaystyle f\left(x+\Delta x\right)

d) \displaystyle f\left(x+1\right)-f\left(x-1\right)

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SOLUTION:

Part A

\begin{align*}
f\left(-5\right) & =\left(-5\right)^2-4\left(-5\right)\\
& =25+20\\
& =45 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

\begin{align*}
f\left(y^2+1\right) & = \left(y^2+1\right)^2-4\left(y^2+1\right)\\
& =y^4+2y^2+1-4y^2-4\\
& =y^4-2y^2-3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

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

Part D

\begin{align*}
f\left(x+1\right)-f\left(x-1\right) & =\left[\left(x+1\right)^2-4\left(x+1\right)\right]-\left[\left(x-1\right)^2-4\left(x-1\right)\right]\\
& = \left[x^2+2x+1-4x-4\right]-\left[x^2-2x+1-4x+4\right]\\
& =x^2-x^2+2x-4x+2x+4x+1-4-1-4\\
& =4x-8\\
& =4\left(x-2\right) \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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