Tag Archives: Feliciano and Uy

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

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PROBLEM:

If y=x2+3x\displaystyle y=\frac{x^2+3}{x}, find xx as a function of yy.


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SOLUTION:

y=x2+3xxy=x2+3x2xy+3=0\begin{align*} y & = \frac{x^2+3}{x} \\ xy & =x^2+3 \\ x^2-xy+3&=0 \end{align*}

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

x=b±b24ac2ax=(y)±(y)24(1)(3)2(1)x=y±y2122  (Answer)\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 f(x)=x24x\displaystyle f\left(x\right)=x^2-4x, find

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

b) f(y2+1)\displaystyle f\left(y^2+1\right)

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

d) f(x+1)f(x1)\displaystyle f\left(x+1\right)-f\left(x-1\right)


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SOLUTION:

Part A

f(5)=(5)24(5)=25+20=45  (Answer)\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

f(y2+1)=(y2+1)24(y2+1)=y4+2y2+14y24=y42y23  (Answer)\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

f(x+Δx)=(x+Δx)24(x+Δx)=(x+Δx)[(x+Δx)4]=(x+Δx)(x+Δx4)  (Answer)\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

f(x+1)f(x1)=[(x+1)24(x+1)][(x1)24(x1)]=[x2+2x+14x4][x22x+14x+4]=x2x2+2x4x+2x+4x+1414=4x8=4(x2)  (Answer)\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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