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

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PROBLEM:

If f(x)=x2−4x\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(x−1)\displaystyle f\left(x+1\right)-f\left(x-1\right)

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SOLUTION:

Part A

f(−5)=(−5)2−4(−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)2−4(y2+1)=y4+2y2+1−4y2−4=y4−2y2−3  (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)2−4(x+Δx)=(x+Δx)[(x+Δx)−4]=(x+Δx)(x+Δx−4)  (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(x−1)=[(x+1)2−4(x+1)]−[(x−1)2−4(x−1)]=[x2+2x+1−4x−4]−[x2−2x+1−4x+4]=x2−x2+2x−4x+2x+4x+1−4−1−4=4x−8=4(x−2)  (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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