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nataly862011 [7]
3 years ago
9

Base: z(x)=cosx Period:180 Maximum:5 Minimum: -4 What are the transformation? Domain and Range? Graph?

Mathematics
1 answer:
garik1379 [7]3 years ago
8 0

Answer:

The transformations needed to obtain the new function are horizontal scaling, vertical scaling and vertical translation. The resultant function is z'(x) = \frac{1}{2}  + \frac{9}{2} \cdot \cos \left(\frac{\pi\cdot x}{90^{\circ}} \right).

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 z (x) = \cos x the base formula, where x is measured in sexagesimal degrees. This expression must be transformed by using the following data:

T = 180^{\circ} (Period)

z_{min} = -4 (Minimum)

z_{max} = 5 (Maximum)

The cosine function is a periodic bounded function that lies between -1 and 1, that is, twice the unit amplitude, and periodicity of 2\pi radians. In addition, the following considerations must be taken into account for transformations:

1) x must be replaced by \frac{2\pi\cdot x}{180^{\circ}}. (Horizontal scaling)

2) The cosine function must be multiplied by a new amplitude (Vertical scaling), which is:

\Delta z = \frac{z_{max}-z_{min}}{2}

\Delta z = \frac{5+4}{2}

\Delta z = \frac{9}{2}

3) Midpoint value must be changed from zero to the midpoint between new minimum and maximum. (Vertical translation)

z_{m} = \frac{z_{min}+z_{max}}{2}

z_{m} = \frac{1}{2}

The new function is:

z'(x) = z_{m} + \Delta z\cdot \cos \left(\frac{2\pi\cdot x}{T} \right)

Given that z_{m} = \frac{1}{2}, \Delta z = \frac{9}{2} and T = 180^{\circ}, the outcome is:

z'(x) = \frac{1}{2}  + \frac{9}{2} \cdot \cos \left(\frac{\pi\cdot x}{90^{\circ}} \right)

The domain of the function is all real numbers and its range is between -4 and 5. The graph is enclosed below as attachment.

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