Calculate the change in the Ar argument of the function y = (x) = 4 x 7 and the change in the demand of the function 4y. when: × € [0; 1];

The notation in your question seems a bit unclear, but I will address the parts that can be interpreted: 1. For the function \( y(x) = 4x^7 \), the "change in the Ar argument" likely refers to the change in the area under the curve or possibly the change in the function's value (which could be interpreted as the derivative) over the interval \( x \in [0, 1] \). 2. For the second part concerning "the change in the demand of the function \( 4y \)", this could refer to the change in the value of \( 4y \) over the same interval \( x \in [0, 1] \), assuming \( y \) is a function of \( x \). To address the first part, the change in the function's value, \( y(x) = 4x^7 \), over the interval from \( x = 0 \) to \( x = 1 \) is simply the difference in the function's value at those two points. Since any value raised to the seventh power is positive for positive \( x \) and zero for \( x = 0 \), the change will be: \[ y(1) - y(0) = 4(1)^7 - 4(0)^7 = 4 - 0 = 4 \] The second part involves calculating the change in \( 4y \) over the same interval. Since \( y(x) = 4x^7 \), then \( 4y = 16x^7 \). The change in \( 4y \) when \( x \) changes from 0 to 1 is: \[ 4y(1) - 4y(0) = 16(1)^7 - 16(0)^7 = 16 - 0 = 16 \] Thus, the change in \( 4y \) over the interval is 16. If by "change in the Ar argument" or "change in the demand" you meant something different, such as the integral of the function over the interval or the derivative, please clarify so I can assist you accordingly.