The problem involves the functions $f(x) = x^3 - x$ and $g(x) = x^2 + a$, where the tangent line of $y = f(x)$ at point $(x_1, f(x_1))$ is also the tangent line of $y = g(x)$. We need to solve the two parts of the problem:

1. When $x_1 = -1$, find $a$.

The slope of the tangent line of $y = f(x)$ at $x = x_1$ is given by the derivative $f'(x)$. So, calculate $f'(x)$: $f'(x) = 3x^2 - 1.$ At $x_1 = -1$: $f'(-1) = 3(-1)^2 - 1 = 3 - 1 = 2.$ Thus, the slope of the tangent line at $x = -1$ is $k = 2$.

The tangent line passes through $(-1, f(-1))$, so calculate $f(-1)$: $f(-1) = (-1)^3 - (-1) = -1 + 1 = 0.$ The equation of the tangent line is: $y - f(-1) = k(x - x_1),$ which simplifies to: $y - 0 = 2(x + 1) \quad \Rightarrow \quad y = 2x + 2.$

Now, this tangent line must also be the tangent line of $g(x)$. Calculate $g'(x)$: $g'(x) = 2x.$ At the point of tangency, the slope $g'(x_1)$ must equal $k = 2$: $2x_1 = 2 \quad \Rightarrow \quad x_1 = 1.$

However, for consistency with $x_1 = -1$, substitute $x = -1$ into $g(x)$: $g(-1) = (-1)^2 + a = 1 + a.$ Since $g(-1)$ must equal the $y$-intercept of the tangent line ($2$): $1 + a = 2 \quad \Rightarrow \quad a = 1.$

2. Find the range of $a$.

For the tangent line of $f(x)$ at any point $x_1$ to also be a tangent line of $g(x)$, the slopes and $y$-coordinates must match. This requires solving the following equations:

1. Slopes: $f'(x_1) = g'(x_1)$,
2. Tangency: $f(x_1) = g(x_1)$.

From $f'(x_1) = g'(x_1)$: $3x_1^2 - 1 = 2x_1 \quad \Rightarrow \quad 3x_1^2 - 2x_1 - 1 = 0.$ Factorize: $(3x_1 + 1)(x_1 - 1) = 0.$ Thus, $x_1 = -\frac{1}{3}$ or $x_1 = 1$.

For each $x_1$, solve $f(x_1) = g(x_1)$:

• At $x_1 = 1$: $f(1) = 1^3 - 1 = 0, \quad g(1) = 1^2 + a = 1 + a.$ Equating: $1 + a = 0 \quad \Rightarrow \quad a = -1.$

• At $x_1 = -\frac{1}{3}$: $f\left(-\frac{1}{3}\right) = \left(-\frac{1}{3}\right)^3 - \left(-\frac{1}{3}\right) = -\frac{1}{27} + \frac{1}{3} = \frac{8}{27}.$ $g\left(-\frac{1}{3}\right) = \left(-\frac{1}{3}\right)^2 + a = \frac{1}{9} + a.$ Equating: $\frac{1}{9} + a = \frac{8}{27}.$ Simplify: $a = \frac{8}{27} - \frac{3}{27} = \frac{5}{27}.$

Thus, the range of $a$ is: $a \in \left[\frac{5}{27}, -1\right].$

Final Answer:

1. $a = 1$ when $x_1 = -1$,
2. $a \in \left[\frac{5}{27}, -1\right]$.

Would you like further clarification or details?

1. How is the tangent line equation derived from a point and slope?
2. Why is solving $f'(x) = g'(x)$ important for tangency?
3. Can you confirm the consistency of slopes at $x_1 = -\frac{1}{3}$?
4. How is the range of $a$ linked to different $x_1$ values?
5. What role does the quadratic factorization play in finding $x_1$?

Tip: Always double-check derivatives and substitutions to avoid minor calculation errors.