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(a) Find all $x$ such that $f(x) \leq 2$ where

$$

f(x)=-x^2+1 \quad f(x)=(x-1)^2 \quad f(x)=x^3

$$

Write your answers in interval notation and draw them on the graphs of the functions.

(b) Using the functions in part a, find all $x$ such that $|f(x)| \leq 2$. Write your answers in interval notation and draw them on the graphs of the functions.

(c) Can you find upper bounds for the functions in part a? That is, for each function $f$ is there a number $M$ such that for all $x, f(x) \leq M$ ?

(d) What about lower bounds for the functions in part a? That is, for each $f$ can you find a number $m$ such that for all $x, f(x) \geq m$ ?

(e) What about finding upper and lower bounds for these functions restricted to the interval $[-1,1]$ ? That is, for each $f$ can you find numbers $M$ and $m$ such that for all $x$ in $[-1,1], m \leq f(x) \leq M ?$

(f) True or false? If $M$ is an upper bound for the function $f$ and $M^{\prime}$ is an upper bound for the function $g$, then for all $x$ which are in the domains of both $f$ and g,

$$

|f(x)+g(x)| \leq M+M^{\prime} .

$$

(a) To find all $x$ such that $f(x) \leq 2$ for each function, we can set up the inequality and solve for $x$.

For the function $f(x)=-x^2+1$, we have $-x^2+1 \leq 2$, which simplifies to $x^2 \geq -1$. Since $x^2$ is always nonnegative, this inequality is satisfied for all $x$. Therefore, the solution is $(-\infty, \infty)$.

For the function $f(x)=(x-1)^2$, we have $(x-1)^2 \leq 2$, which simplifies to $(x-1)^2-2 \leq 0$. By solving this inequality, we find $-1 \leq x \leq 3$. Therefore, the solution is $[-1, 3]$.

For the function $f(x)=x^3$, we have $x^3 \leq 2$. Since the cube of any real number can be either positive or negative, there are no real solutions to this inequality. Therefore, the solution is the empty set, $\emptyset$.

The graphs of these functions can be drawn accordingly to visualize the solutions.

(b) To find all $x$ such that $|f(x)| \leq 2$ for each function, we consider the absolute value of each function.

For $f(x)=-x^2+1$, we have $|f(x)| = |-x^2+1|$. This absolute value expression is always nonnegative, so $|f(x)| \leq 2$ is satisfied for all $x$. The solution is $(-\infty, \infty)$.

For $f(x)=(x-1)^2$, we have $|f(x)| = |(x-1)^2| = (x-1)^2$. From part (a), we know that the solution to $(x-1)^2 \leq 2$ is $[-1, 3]$. Therefore, the solution to $|f(x)| \leq 2$ is $[-1, 3]$.

For $f(x)=x^3$, we have $|f(x)| = |x^3| = x^3$. Since the cube of any real number can be positive or negative, $|f(x)| \leq 2$ is not satisfied for any $x$. The solution is the empty set, $\emptyset$.

(c) Yes, we can find upper bounds for each function in part (a).

For $f(x)=-x^2+1$, we can see that the maximum value of $f(x)$ is $1$. Therefore, for all $x$, $f(x) \leq 1$.

For $f(x)=(x-1)^2$, the function is always nonnegative or zero. The maximum value is $0$, so for all $x$, $f(x) \leq 0$.

For $f(x)=x^3$, there is no upper bound. The function can take arbitrarily large positive or negative values as $x$ approaches infinity or negative infinity.

(d) For each function in part (a), we can find a lower bound.

For $f(x)=-x^2+1$, we can see that the minimum value of $f(x)$ is $-1$. Therefore, for all $x$, $f(x) \geq -1$.

For $f(x)=(x-1)^2$, the function is always nonnegative or zero. The minimum value is $

(b) Suppose $f(x)=x^2$ and $g(x)=\sin x$.

i. Write the functions in part a in terms of $f$ and $g$. (For example, if $h(x)=2 x^2$ you can write $h$ in terms of $f$ as $h(x)=2 f(x)$.) If you find more than one way of writing these functions in terms of $f$ and $g$, show that they are equivalent.

ii. How can you change the graph of $f$ to obtain the graphs of the first three functions? Use your work from part a to help you.

i. Writing the functions in part (a) in terms of $f$ and $g$:

The first function in part (a) is $f(x)=-x^2+1$. We can write this function in terms of $f$ and $g$ as $f(x) = -f(x) + 2f(x) + 1$. Here, $f(x) = -x^2$ and $g(x) = 1$.

The second function in part (a) is $f(x)=(x-1)^2$. We can write this function in terms of $f$ and $g$ as $f(x) = f(x) + g(x) – 1$. Here, $f(x) = x^2$ and $g(x) = -2x + 1$.

The third function in part (a) is $f(x)=x^3$. We can write this function in terms of $f$ and $g$ as $f(x) = f(x) – 3g(x)$. Here, $f(x) = x^3$ and $g(x) = -x^2$.

These alternative representations are equivalent to the original functions given in part (a).

ii. To change the graph of $f(x)=x^2$ to obtain the graphs of the first three functions, we can use the work from part (a):

For the first function $f(x)=-x^2+1$, we shift the graph of $f(x)$ downward by $1$ unit. This is equivalent to subtracting $1$ from $f(x)$.

For the second function $f(x)=(x-1)^2$, we shift the graph of $f(x)$ to the right by $1$ unit. This is equivalent to replacing $x$ with $(x-1)$ in $f(x)$.

For the third function $f(x)=x^3$, we can see that it already corresponds to the graph of $f(x)$.

By making these adjustments to the graph of $f(x)=x^2$, we obtain the graphs of the first three functions given in part (a).

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