1. **Problem statement:** Given the family of functions $$f_k(x) = \frac{1}{2k} x^2 (x - 2k)^2$$ with parameter $$k > 0$$, we need to analyze the roots, the distance between extrema, and the area enclosed by the graph, the x-axis, and vertical lines $$x = -1$$ and $$x = 1$$. 2. **Finding the roots:** To find the zeros of $$f_k(x)$$, set $$f_k(x) = 0$$: $$\frac{1}{2k} x^2 (x - 2k)^2 = 0$$ Since $$\frac{1}{2k} \neq 0$$ for $$k > 0$$, the roots come from: $$x^2 = 0 \quad \Rightarrow \quad x = 0$$ and $$(x - 2k)^2 = 0 \quad \Rightarrow \quad x = 2k$$ Thus, the function has exactly two roots at $$x = 0$$ and $$x = 2k$$. 3. **Distance between the local maximum and the two local minima:** First, find the critical points by differentiating $$f_k(x)$$: $$f_k(x) = \frac{1}{2k} x^2 (x - 2k)^2$$ Rewrite: $$f_k(x) = \frac{1}{2k} x^2 (x^2 - 4kx + 4k^2) = \frac{1}{2k} (x^4 - 4kx^3 + 4k^2 x^2)$$ Differentiate: $$f_k'(x) = \frac{1}{2k} (4x^3 - 12kx^2 + 8k^2 x) = \frac{1}{2k} x (4x^2 - 12kx + 8k^2)$$ Set derivative to zero: $$f_k'(x) = 0 \Rightarrow x = 0 \quad \text{or} \quad 4x^2 - 12kx + 8k^2 = 0$$ Solve quadratic: $$4x^2 - 12kx + 8k^2 = 0$$ Divide by 4: $$x^2 - 3kx + 2k^2 = 0$$ Use quadratic formula: $$x = \frac{3k \pm \sqrt{9k^2 - 8k^2}}{2} = \frac{3k \pm k}{2}$$ So, $$x_1 = \frac{3k - k}{2} = k$$ $$x_2 = \frac{3k + k}{2} = 2k$$ Critical points are at $$x = 0, k, 2k$$. Evaluate $$f_k(x)$$ at these points: - At $$x=0$$: $$f_k(0) = 0$$ - At $$x=k$$: $$f_k(k) = \frac{1}{2k} k^2 (k - 2k)^2 = \frac{1}{2k} k^2 (-k)^2 = \frac{1}{2k} k^2 k^2 = \frac{k^3}{2}$$ - At $$x=2k$$: $$f_k(2k) = 0$$ Since $$f_k(k) > 0$$ and $$f_k(0) = f_k(2k) = 0$$, the point at $$x=k$$ is a local maximum, and points at $$x=0$$ and $$x=2k$$ are local minima (actually roots). The problem states the local maximum has the same distance to the two local minima. The distance between $$x=k$$ and $$x=0$$ is $$k$$, and between $$x=k$$ and $$x=2k$$ is also $$k$$. Therefore, the distance is $$k$$. 4. **Area enclosed by $$G_k$$, the x-axis, and lines $$x=-1$$ and $$x=1$$:** The integral $$\int_{-1}^1 f_k(x) dx$$ gives the net area between the curve and the x-axis over $$[-1,1]$$. Since $$f_k(x) \geq 0$$ for all $$x$$ (because it is a product of squares and positive constants), the graph lies above or on the x-axis. Therefore, the integral $$\int_{-1}^1 f_k(x) dx$$ equals the area enclosed by $$G_k$$, the x-axis, and the vertical lines $$x=-1$$ and $$x=1$$. Hence, the statement is **true**. **Final answers:** - Roots: $$x=0$$ and $$x=2k$$ - Distance between local maximum and local minima: $$k$$ - The integral $$\int_{-1}^1 f_k(x) dx$$ gives the area enclosed by the graph, x-axis, and lines $$x=-1$$ and $$x=1$$.