Answer:
10'-13 coordinates make of this abscissa
Answer:
Take a look at the 'proof' below
Step-by-step explanation:
The questions asks us to determine the anti-derivative of the function f(x) = 4x^3 sec^2 x^4. Let's start by converting this function into integral form. That would be the following:
Now all we have to do is solve the integral. Let's substitute 'u = x^4' into the equation 'du/dx = 4x^3.' We will receive dx = 1/4x^3 du. If we simplify a bit further:
Our hint tells us that d/dx tan(x) = sec^2(x). Similarly in this case our integral boils down to tan(u). If we undo the substitution, we will receive the expression tan(x^4). Therefore you are right, the first option is an anti-derivative of the function f(x) = 4x^3 sec^2 x^4.
9514 1404 393
Answer:
Step-by-step explanation:
The extrema will be at the ends of the interval or at a critical point within the interval.
The derivative of the function is ...
f'(x) = 3x² -4x -4 = (x -2)(3x +2)
It is zero at x=-2/3 and at x=2. Only the latter critical point is in the interval. Since the leading coefficient of this cubic is positive, the right-most critical point is a local minimum. The coordinates of interest in this interval are ...
f(0) = 2
f(2) = ((2 -2)(2) -4)(2) +2 = -8 +2 = -6
f(3) = ((3 -2)(3) -4)(3) +2 = -3 +2 = -1
The absolute maximum on the interval is f(0) = 2.
The absolute minimum on the interval is f(2) = -6.
Step-by-step explanation:
the answer is D and is that all ur work?
Answer:
X =
Y = 204 =
Step-by-step explanation: