1. **State the problem:** We are given two functions $f(x) = x^2 + kx$ and $g(x) = \sqrt{x} + 1$. We need to find all values of $k$ such that the compositions satisfy $$(f \circ g)(x) = (g \circ f)(x)$$ for all $x$ in the domain where both sides are defined. 2. **Write the compositions:** - $$(f \circ g)(x) = f(g(x)) = f(\sqrt{x} + 1) = (\sqrt{x} + 1)^2 + k(\sqrt{x} + 1)$$ - $$(g \circ f)(x) = g(f(x)) = g(x^2 + kx) = \sqrt{x^2 + kx} + 1$$ 3. **Expand and simplify $(f \circ g)(x)$:** $$ (\sqrt{x} + 1)^2 + k(\sqrt{x} + 1) = (\sqrt{x})^2 + 2\sqrt{x} + 1 + k\sqrt{x} + k = x + 2\sqrt{x} + 1 + k\sqrt{x} + k $$ Combine like terms: $$ x + (2 + k)\sqrt{x} + (1 + k) $$ 4. **Set the equality:** $$ x + (2 + k)\sqrt{x} + (1 + k) = \sqrt{x^2 + kx} + 1 $$ 5. **Isolate the square root on the right:** $$ \sqrt{x^2 + kx} = x + (2 + k)\sqrt{x} + (1 + k) - 1 = x + (2 + k)\sqrt{x} + k $$ 6. **Square both sides to eliminate the square root:** $$ x^2 + kx = \left(x + (2 + k)\sqrt{x} + k\right)^2 $$ 7. **Expand the right side:** Let $a = x$, $b = (2 + k)\sqrt{x}$, and $c = k$. $$ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc $$ Calculate each term: - $a^2 = x^2$ - $b^2 = ((2 + k)\sqrt{x})^2 = (2 + k)^2 x$ - $c^2 = k^2$ - $2ab = 2 \cdot x \cdot (2 + k)\sqrt{x} = 2(2 + k) x^{3/2}$ - $2ac = 2 \cdot x \cdot k = 2kx$ - $2bc = 2 \cdot (2 + k)\sqrt{x} \cdot k = 2k(2 + k) \sqrt{x}$ So, $$ x^2 + kx = x^2 + (2 + k)^2 x + k^2 + 2(2 + k) x^{3/2} + 2kx + 2k(2 + k) \sqrt{x} $$ 8. **Subtract $x^2$ from both sides:** $$ kx = (2 + k)^2 x + k^2 + 2(2 + k) x^{3/2} + 2kx + 2k(2 + k) \sqrt{x} $$ 9. **Bring all terms to one side:** $$ 0 = (2 + k)^2 x + k^2 + 2(2 + k) x^{3/2} + 2kx + 2k(2 + k) \sqrt{x} - kx $$ Group like terms: $$ 0 = (2 + k)^2 x + 2kx - kx + k^2 + 2(2 + k) x^{3/2} + 2k(2 + k) \sqrt{x} $$ Simplify $x$ terms: $$ 0 = (2 + k)^2 x + (2k - k) x + k^2 + 2(2 + k) x^{3/2} + 2k(2 + k) \sqrt{x} $$ $$ 0 = (2 + k)^2 x + k x + k^2 + 2(2 + k) x^{3/2} + 2k(2 + k) \sqrt{x} $$ 10. **Rewrite the equation grouping powers of $x$:** $$ 0 = k^2 + \left((2 + k)^2 + k\right) x + 2k(2 + k) \sqrt{x} + 2(2 + k) x^{3/2} $$ 11. **For the equality to hold for all $x \geq 0$ (domain of $g$), the coefficients of each power of $x$ must be zero:** - Constant term: $$k^2 = 0 \implies k = 0$$ - Coefficient of $x$: $$ (2 + k)^2 + k = 0 $$ - Coefficient of $\sqrt{x}$: $$ 2k(2 + k) = 0 $$ - Coefficient of $x^{3/2}$: $$ 2(2 + k) = 0 $$ 12. **Analyze each condition:** - From $k^2 = 0$, we get $k = 0$. - From $2(2 + k) = 0$, we get $2 + k = 0 \implies k = -2$. - From $2k(2 + k) = 0$, either $k = 0$ or $k = -2$. - From $(2 + k)^2 + k = 0$, substitute $k = 0$: $$ (2 + 0)^2 + 0 = 4 \neq 0 $$ Substitute $k = -2$: $$ (2 - 2)^2 + (-2) = 0 + (-2) = -2 \neq 0 $$ 13. **Conclusion:** No single $k$ satisfies all conditions simultaneously for all $x$. However, check if $k=0$ or $k=-2$ satisfy the original equality for some $x$. - For $k=0$: $$ f(x) = x^2 $$ $$ (f \circ g)(x) = (\sqrt{x} + 1)^2 = x + 2\sqrt{x} + 1 $$ $$ (g \circ f)(x) = \sqrt{x^2} + 1 = |x| + 1 $$ For $x \geq 0$, $|x| = x$, so: $$ (g \circ f)(x) = x + 1 $$ These are equal only if $2\sqrt{x} = 0 \implies x=0$. - For $k=-2$: $$ f(x) = x^2 - 2x $$ $$ (f \circ g)(x) = (\sqrt{x} + 1)^2 - 2(\sqrt{x} + 1) = x + 2\sqrt{x} + 1 - 2\sqrt{x} - 2 = x - 1 $$ $$ (g \circ f)(x) = \sqrt{x^2 - 2x} + 1 $$ For $x=1$, $$ (f \circ g)(1) = 1 - 1 = 0 $$ $$ (g \circ f)(1) = \sqrt{1 - 2} + 1 = \sqrt{-1} + 1 $$ Not real, so no equality. 14. **Final answer:** The only value of $k$ for which $(f \circ g)(x) = (g \circ f)(x)$ holds for all $x$ in the domain is **none**. The equality holds only at specific points for $k=0$ or $k=-2$, but not for all $x$. **Summary:** There is no $k$ such that the compositions are equal for all $x$ in the domain.