Complete the following statement.
1 answer:
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The maxima of f(x) occur at its critical points, where f '(x) is zero or undefined. We're given f '(x) is continuous, so we only care about the first case. Looking at the plot, we see that f '(x) = 0 when x = -4, x = 0, and x = 5.
Notice that f '(x) ≥ 0 for all x in the interval [0, 5]. This means f(x) is strictly increasing, and so the absolute maximum of f(x) over [0, 5] occurs at x = 5.
By the fundamental theorem of calculus,
The definite integral corresponds to the area of a trapezoid with height 2 and "bases" of length 5 and 2, so
Answer:
<h2>X=6</h2>Step-by-step explanation:
<h3>to understand this</h3><h3>you need to know about:</h3>- equation
- PEMDAS
Givens
y = 2
x = 1
z(the hypotenuse) = √(2^2 + 1^2) = √5
Cos(u) = x value / hypotenuse = 1/√5
Sin(u) = y value / hypotenuse = 2/√5
Solve for sin2u
Sin(2u) = 2*sin(u)*cos(u)
Sin(2u) = 2() = 4/5
Solve for cos(2u)
cos(2u) = - sqrt(1 - sin^2(2u))
Cos(2u) = - sqrt(1 - (4/5)^2 )
Cos(2u) = -sqrt(1 - 16/25)
cos(2u) = -sqrt(9/25)
cos(2u) = -3/5
Solve for Tan(2u)
tan(2u) = sin(2u) / cos(2u) = 4/5// - 3/5 = - 0.8/0.6 = - 1.3333 = - 4/3
Notes
One: Notice that you would normally rationalize the denominator, but you don't have to in this case. The formulas are such that they perform the rationalizations themselves.
Two: Notice the sign on the cos(2u). The sin is plus even though the angle (2u) is in the second quadrant. The cos is different. It is about 126 degrees which would make it a negative root (9/25)
Three: If you are uncomfortable with the tan, you could do fractions.