Fonksiyon ve Türevi Arasındaki Grafiksel İlişki

Here we have a problem involving the graphs of a function f of x and its derivative f prime of x. The coordinate axes have been deleted, but we have a grid of equal squares. We need to find the minimum value of f of x.

Let's carefully analyze the grid points marked with dots. Notice the horizontal distance between the leftmost dots and the intersection point. Let's count the grid squares.

Between the left column of dots and the intersection point on the right, we span exactly 2 horizontal grid units. Let's verify the vertical rise of the line graph over this distance.

The line rises 3 squares for every 2 squares it moves to the right. This means the slope of the derivative graph is 3 over 2, or 1.5.

Since the slope of the $f'(x)$ graph represents the second derivative, we know that $f''(x) = 1.5$. For a quadratic $f(x) = ax^2 + bx + c$, the second derivative is $2a$.

Now let's find the location of the vertex. Notice that at the left point (x=0 relative), the parabola is 3 units above the line start. At the right point (x=2 relative), they intersect, so they are at the same height.

Actually, the visual symmetry is simpler. The parabola height relative to the line at the left is 3 units. At the intersection, the height relative to the line is 0. But more importantly, look at the heights relative to the grid.

Comparing the start point and intersection point: The parabola dot on the left is exactly horizontal with the intersection dot on the right. This means $f(0) = f(2)$ in our local coordinates.

Since the parabola has the same height at $x=0$ and $x=2$ (relative to grid), the vertex must be exactly in the middle, at local $x=1$.