The greatest length of the considered pendent is given by: Option A: 1.25 cm
Suppose the considered function is [tex]y = f(x)[/tex]
Then, we will plot points as [tex](x,y) =(x,f (x))[/tex] for all values of 'x' which are in domain of the function.
For this case, the given function's graph (the graph of [tex]y = \dfrac{x^2}{20}[/tex] ) is shown below.
We're given that the largest dimension is of 10 units.
Let this be the line segment parallel to the x-axis, intersecting y-axis at [tex]y = y_0[/tex]
Now, since the curve is symmetric over (as for x or -x, we get same y), so the line segment that is parallel to x axis also intersects the graph at symmetric places. Let those points be in front of [tex]x_0[/tex] and [tex]-x_0[/tex]
For those inputs, the output is [tex]y = y_0[/tex],
Also, we have length of the line segment as of 10 units, this is the difference between those symmetric inputs.
Thus, we get two equations:
[tex]x_0 - (-x_0) = 10[/tex]
[tex]f(x_0) = y_0 = \dfrac{x_0^2}{20}[/tex]
From first one, we get:
[tex]x_0 = 10/2= 5[/tex]
Thus, we get:
[tex]f(x_0) = y_0 = \dfrac{x_0^2}{20}\\\\f(5) = y_0=\dfrac{25}{20} = 1.25[/tex]
Since the length of y_0 is the shortest distance from x-axis to that parallel line segment, this is the second dimension needed.
Thus, the greatest length of the considered pendent is given by: Option A: 1.25 cm
Learn more about functions here:
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