The exercise provided has the function $f(x) = \frac{x^2}{x + 1}$ and asks to evaluate several expressions based on this function. Let's go through each part:

Given Function:

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

Tasks:

Let's evaluate each expression, one by one.

(a) $f(-3)$

To find $f(-3)$, substitute $x = -3$ into $f(x)$: $f(-3) = \frac{(-3)^2}{-3 + 1} = \frac{9}{-2} = -\frac{9}{2}$

(b) $f(1)$

To find $f(1)$, substitute $x = 1$ into $f(x)$: $f(1) = \frac{1^2}{1 + 1} = \frac{1}{2}$

(c) $f(0)$

To find $f(0)$, substitute $x = 0$ into $f(x)$: $f(0) = \frac{0^2}{0 + 1} = 0$

(d) $f(t)$

Since $f(t)$ just involves substituting $t$ for $x$, we have: $f(t) = \frac{t^2}{t + 1}$

(e) $f(-x)$

To find $f(-x)$, substitute $x = -x$ into $f(x)$: $f(-x) = \frac{(-x)^2}{-x + 1} = \frac{x^2}{-x + 1}$

(f) $-f(x)$

To find $-f(x)$, take the negative of $f(x)$: $-f(x) = -\frac{x^2}{x + 1}$

(g) $f(2x)$

To find $f(2x)$, substitute $x = 2x$ into $f(x)$: $f(2x) = \frac{(2x)^2}{2x + 1} = \frac{4x^2}{2x + 1}$

(h) $f(x - 2)$

To find $f(x - 2)$, substitute $x = x - 2$ into $f(x)$: $f(x - 2) = \frac{(x - 2)^2}{(x - 2) + 1} = \frac{x^2 - 4x + 4}{x - 1}$

(i) $f(x + h)$

To find $f(x + h)$, substitute $x = x + h$ into $f(x)$: $f(x + h) = \frac{(x + h)^2}{(x + h) + 1} = \frac{x^2 + 2xh + h^2}{x + h + 1}$

Would you like further explanations on any of these steps?

Related Questions:

1. How does the behavior of $f(x)$ change as $x$ approaches infinity?
2. What is the domain and range of $f(x) = \frac{x^2}{x+1}$?
3. How would you calculate the derivative of $f(x)$?
4. What are the asymptotes of $f(x)$ if any exist?
5. How would the expression change if we use a different function, such as $g(x) = \frac{x^3}{x+2}$?

Tip:

When substituting expressions, always simplify carefully to avoid mistakes with negative signs and terms in the denominator.