1. **Problem statement:** We are given the function $$y = \frac{1}{2x^2} + 3x$$ for $$-1 \leq x \leq 3$$. (i) Estimate the gradient of the curve at $$x=0.5$$ by drawing a tangent. (ii) Estimate the solution(s) of the equation $$\frac{1}{2x^2} + 3x = 2$$ using the graph. (iii) Estimate the solution(s) of the equation $$\frac{1}{2x^2} = 7 - 4x$$ by drawing a suitable line on the grid. --- 2. **Formula and rules:** - The gradient of a curve at a point is the slope of the tangent line at that point. - To find the gradient at $$x=0.5$$, we estimate the slope of the tangent line drawn at that point. - To solve equations graphically, find the x-values where the curve intersects the horizontal line representing the constant on the right side. --- 3. **Step (i): Estimate gradient at $$x=0.5$$** - Draw a tangent line at $$x=0.5$$ on the graph. - Estimate two points on the tangent line, say $$(x_1,y_1)$$ and $$(x_2,y_2)$$. - Calculate gradient $$m = \frac{y_2 - y_1}{x_2 - x_1}$$. - From the graph, the tangent appears to rise about 4 units when moving 1 unit right, so gradient $$\approx 4$$. --- 4. **Step (ii): Estimate solution of $$\frac{1}{2x^2} + 3x = 2$$** - On the graph, find where $$y = \frac{1}{2x^2} + 3x$$ intersects the horizontal line $$y=2$$. - From the graph, this occurs near $$x \approx 0.3$$. --- 5. **Step (iii): Estimate solutions of $$\frac{1}{2x^2} = 7 - 4x$$** - Rewrite as $$y_1 = \frac{1}{2x^2}$$ and $$y_2 = 7 - 4x$$. - Draw the line $$y = 7 - 4x$$ on the graph. - Find intersection points of $$y_1$$ and $$y_2$$. - From the graph, intersections appear near $$x \approx 0.5$$ and $$x \approx 1.5$$. --- **Final answers:** (i) Gradient at $$x=0.5$$ is approximately $$4$$. (ii) Solution to $$\frac{1}{2x^2} + 3x = 2$$ is approximately $$x=0.3$$. (iii) Solutions to $$\frac{1}{2x^2} = 7 - 4x$$ are approximately $$x=0.5$$ and $$x=1.5$$.