The derivative of the function g(x) as given in the task content by virtue of the Fundamental theorem of calculus is; g'(x) = √2 ln(t) dt = 1.
<h3>What is the derivative of the function g(x) by virtue of the Fundamental theorem of calculus as given in the task content?</h3>
g(x) = Integral; √2 ln(t) dt (with the upper and lower limits e^x and 1 respectively).
Since, it follows from the Fundamental theorem of calculus that given an integral where;
Now, g(x) = Integral f(t) dt with limits a and x, it follows that the differential of g(x);
g'(x) = f(x).
Consequently, the function g'(x) which is to be evaluated in this scenario can be determined as:
g'(x) =
= 1
The derivative of the function g(x) as given in the task content by virtue of the Fundamental theorem of calculus is; g'(x) = √2 ln(t) dt = 1.
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The first digit is the largest, the second is NOT the smallest
The third digit is NOT even and the first digit is odd.
Part A:
Function A:
Slope = (7-3)/(5-3) = 4/2 = 2
Equation:
y - 3 = 2(x - 3)
y - 3 = 2x - 6
y = 2x - 3
Funcion B:
(0, 3 ) and (-5, 0)
Slope = (3 - 0)/(0 + 5) = 3/5
y-intercept (0,3) so b = 3
Equation:
y = 3/5 x + 3
Function C:
y = 3x + 1
Part B:
Rate of change is the change in y over the change in x (rise/run). It's also the slope
Function A: rate of change = 2
Function B: rate of change = 3/5 (smallest)
Function C: rate of change = 3 (largest)
Order linear functions based on rate of change from least to greatest.
Function B: y = 3/5 x + 3
Function A: y = 2x - 3
Function C: y = 3x + 1
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