Answer:
The transformations needed to obtain the new function are horizontal scaling, vertical scaling and vertical translation. The resultant function is [tex]z'(x) = \frac{1}{2} + \frac{9}{2} \cdot \cos \left(\frac{\pi\cdot x}{90^{\circ}} \right)[/tex].
The domain of the function is all real numbers and its range is between -4 and 5.
The graph is enclosed below as attachment.
Step-by-step explanation:
Let be [tex]z (x) = \cos x[/tex] the base formula, where [tex]x[/tex] is measured in sexagesimal degrees. This expression must be transformed by using the following data:
[tex]T = 180^{\circ}[/tex] (Period)
[tex]z_{min} = -4[/tex] (Minimum)
[tex]z_{max} = 5[/tex] (Maximum)
The cosine function is a periodic bounded function that lies between -1 and 1, that is, twice the unit amplitude, and periodicity of [tex]2\pi[/tex] radians. In addition, the following considerations must be taken into account for transformations:
1) [tex]x[/tex] must be replaced by [tex]\frac{2\pi\cdot x}{180^{\circ}}[/tex]. (Horizontal scaling)
2) The cosine function must be multiplied by a new amplitude (Vertical scaling), which is:
[tex]\Delta z = \frac{z_{max}-z_{min}}{2}[/tex]
[tex]\Delta z = \frac{5+4}{2}[/tex]
[tex]\Delta z = \frac{9}{2}[/tex]
3) Midpoint value must be changed from zero to the midpoint between new minimum and maximum. (Vertical translation)
[tex]z_{m} = \frac{z_{min}+z_{max}}{2}[/tex]
[tex]z_{m} = \frac{1}{2}[/tex]
The new function is:
[tex]z'(x) = z_{m} + \Delta z\cdot \cos \left(\frac{2\pi\cdot x}{T} \right)[/tex]
Given that [tex]z_{m} = \frac{1}{2}[/tex], [tex]\Delta z = \frac{9}{2}[/tex] and [tex]T = 180^{\circ}[/tex], the outcome is:
[tex]z'(x) = \frac{1}{2} + \frac{9}{2} \cdot \cos \left(\frac{\pi\cdot x}{90^{\circ}} \right)[/tex]
The domain of the function is all real numbers and its range is between -4 and 5. The graph is enclosed below as attachment.
