Tag Archives: differential calculus solutions

Differential and Integral Calculus by Feliciano and Uy Banner

Differential and Integral Calculus by Feliciano and Uy Complete Solution Manual



Advertisements
Advertisements

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1, Problem 2

Advertisements
Advertisements

PROBLEM:

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


Advertisements
Advertisements

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*}

Advertisements
Advertisements

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1 Problem 1

Advertisements
Advertisements

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)


Advertisements
Advertisements

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*}

Advertisements
Advertisements