Tag Archives: Limits

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 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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