Questions: For each skeleton below, you will select the most fitting derivative, considering only the sign of (f^prime(x)) on the interval ((-1.5,1.5)^*); from the drop-down menu. This corresponds to a function (f(x)), whose derivative is [ f^prime(x)= 1+x^2 sqrt1-x^2 (1+x)^2 (1-x)^2 x^2-1 ]

Solution Steps

The graph of $f(x)$ is increasing for $x<-1$. This means $f'(x)>0$ for $x<-1$. The graph is constant for $-1<x<1$, indicating $f'(x)=0$ for $-1<x<1$. Lastly, the graph is increasing for $x>1$, which implies $f'(x)>0$ for $x>1$.

We are looking for a function that is positive for $x<-1$, zero for $-1<x<1$, and positive for $x>1$. This means $f'(x)$ should be zero only for $x=\pm1$. Let us consider the interval (-1.5, 1.5).

1. $1+x^2$: This option is always positive in the interval, hence it does not fit the given graph.
2. $\sqrt{1-x^2}$: This option is positive for $-1<x<1$ and zero for $x=\pm1$. Outside this interval, the function is not defined in the real domain.
3. $(1+x)^2$: This option is positive everywhere except at $x=-1$, where it is zero. This does not satisfy the required condition of being zero in the interval $(-1, 1)$.
4. $(1-x)^2$: This option is positive everywhere except at $x=1$, where it is zero. Similar to the previous option, this also doesn't satisfy the required condition.
5. $x^2-1$: This option is negative for $-1<x<1$ and zero for $x=\pm1$. It is positive for $|x|>1$. Although it is negative for $-1<x<1$, when considering just the sign, the absolute value $|x^2-1|$ is positive for $|x|>1$ and 0 for $x=\pm1$, matching the given graph's behavior.

Final Answer: The final answer is $\boxed{x^2-1}$