In Problems 39– 46, find the following for each function:

$\left(a\right)f\left(0\right)\left(b\right)f\left(1\right)\left(c\right)f(-1)\left(d\right)f(-x)\left(e\right)-f\left(x\right)\left(f\right)f(x+1)\left(g\right)f\left(x\right)\left(h\right)f(x+h)$

$f\left(x\right)=1-\frac{1}{{(x+2)}^2}$

We have to find the value for each function:

$\left(a\right)f\left(0\right)\left(b\right)f\left(1\right)\left(c\right)f(-1)\left(d\right)f(-x)\left(e\right)-f\left(x\right)\left(f\right)f(x+1)\left(g\right)f\left(x\right)\left(h\right)f(x+h)$

The given equation is$f\left(x\right)=1-\frac{1}{{(x+2)}^2}$

$$\begin{gathered} f\left(0\right)=1-\frac{1}{{(0+2)}^2} \\ f\left(0\right)=1-\frac{1}{{\left(2\right)}^2} \\ f\left(0\right)=1-\frac{1}{4} \\ f\left(0\right)=1\frac{4}{4}-\frac{1}{4} \\ f\left(0\right)=\frac{4-1}{4} \\ f\left(0\right)=\frac{3}{4} \end{gathered}$$

$$\begin{gathered} f\left(1\right)=1-\frac{1}{{(1+2)}^2} \\ f\left(1\right)=1-\frac{1}{{\left(3\right)}^2} \\ f\left(1\right)=1-\frac{1}{9} \\ f\left(1\right)=1\times \frac{9}{9}-\frac{1}{9} \\ f\left(1\right)=\frac{9-1}{9} \\ f\left(1\right)=\frac{8}{9} \end{gathered}$$

$$\begin{gathered} f(-1)=1-\frac{1}{{(-1+2)}^2} \\ f(-1)=1-\frac{1}{{\left(1\right)}^2} \\ f(-1)=1-\frac{1}{1} \\ f(-1)=1-1 \\ f(-1)=0 \end{gathered}$$

$f(-x)=1-\frac{1}{{(-x+2)}^2}$

$$\begin{gathered} f(-x)=1\times \frac{{(-x+2)}^2}{{(-x+2)}^2}-\frac{1}{{(-x+2)}^2} \\ f(-x)=\frac{{(-x+2)}^2-1}{{(-x+2)}^2} \end{gathered}$$

Using the formula${(a+b)}^2=a^2+2ab+b^2$

role="math" localid="1645964203501" $$\begin{gathered} f(-x)=\frac{{(-x)}^2+2\times (-x)\times 2+{\left(2\right)}^2-1}{{(-x)}^2+2\times (-x)\times 2+{\left(2\right)}^2} \\ f(-x)=\frac{x^2-4x+4-1}{x^2-4x+4} \\ f(-x)=\frac{x^2-4x+3}{x^2-4x+4} \end{gathered}$$

$$\begin{gathered} -f\left(x\right)=-\left(1-\frac{1}{{(x+2)}^2}\right) \\ -f\left(x\right)=\frac{1}{{(x+2)}^2}-1 \\ -f\left(x\right)=\frac{1}{{(-x+2)}^2}-1\times \frac{{(-x+2)}^2}{{(-x+2)}^2} \\ -f\left(x\right)=\frac{1-{(-x+2)}^2}{{(-x+2)}^2} \end{gathered}$$

Using the formula${(a+b)}^2=a^2+2ab+b^2$

role="math" localid="1645964175112" $$\begin{gathered} -f\left(x\right)=\frac{1-\left\{{(-x)}^2+2\times (-x)\times 2+{\left(2\right)}^2\right\}}{{(-x)}^2+2\times (-x)\times 2+{\left(2\right)}^2} \\ -f\left(x\right)=\frac{1-\left(x^2-4x+4\right)}{x^2-4x+4} \\ -f\left(x\right)=\frac{1-x^2+4x-4}{x^2-4x+4} \\ -f\left(x\right)=\frac{-x^2+4x-3}{x^2-4x+4} \end{gathered}$$

$$\begin{gathered} f(x+1)=1-\frac{1}{{(x+1+2)}^2} \\ f(x+1)=1\times \frac{{(x+3)}^2}{{(x+3)}^2}-\frac{1}{{(x+3)}^2} \\ f(x+1)=\frac{{(x+3)}^2-1}{{(x+3)}^2} \end{gathered}$$

Using the formula${(a+b)}^2=a^2+2ab+b^2$

$$\begin{gathered} f(x+1)=\frac{x^2+6x+9-1}{x^2+6x+9} \\ f(x+1)=\frac{x^2+6x+8}{x^2+6x+9} \end{gathered}$$

The value of$f\left(x\right)=1-\frac{1}{{(x+2)}^2}$

$$\begin{gathered} f(x+h)=1-\frac{1}{{(x+h)}^2} \\ f(x+h)=1\times \frac{{(x+h)}^2}{{(x+h)}^2}-\frac{1}{{(x+h)}^2} \\ f(x+h)=\frac{{(x+h)}^2-1}{{(x+h)}^2} \end{gathered}$$

Using the formula${(a+b)}^2=a^2+2ab+b^2$

$f(x+h)=\frac{{x^2+2xh+h}^2-1}{{x^2+2xh+h}^2}$