Given f(x)=cos(5x)on the closed interval [0,2π/5]
Find the open intervals where f is concave down and the open intervals where f is concave up. Express your answers in interval notation with exact values.
f"(x) = -25cos(5x)
f is concave up where f" is positive: (π/10,3π/10)
f is concave down on [0,π/10)U(3π/10,4π/10]
see the graph at
https://www.wolframalpha.com/input/?i=cos%285x%29
but what would it be within the interval though?
f' = -5 sin 5x
f" = -25 cos 5x
where f" is negative, concave down (sheds water)
where f" is positive, concave up (holds water)
f" is negative where cos 5x is positive
that is where 5x = 0 to 5x = pi/2 and from 5x = 3 pi/2 to 5x = 2 pi
the rest of the circle is where f" is positive
what are the critical values for this problem?
as always, where f'(x) = 0
f' = -5sin(5x)
where is sin(5x)=0 ?
Verify your answer with the graph referenced above.
For heavens sake graph it !
To determine where a function is concave up or concave down, we need to find the second derivative of the function and analyze its sign.
Let's start by finding the first derivative of f(x).
f(x) = cos(5x)
f'(x) = d/dx [cos(5x)]
To differentiate cos(5x), we can use the chain rule. The derivative of the outer function cos(u) is -sin(u), and the derivative of the inner function 5x is 5.
f'(x) = -5sin(5x)
Now, to find the second derivative f''(x), we differentiate the first derivative.
f''(x) = d/dx[-5sin(5x)]
Using the chain rule again, the derivative of -5sin(5x) is -5(cos(5x))(5) = -25cos(5x).
Now, we need to analyze the sign of f''(x) to determine where f(x) is concave up or down.
To find where f''(x) is positive (indicating concave up), we solve the inequality:
-25cos(5x) > 0
Dividing both sides by -25 (a negative number) changes the direction of the inequality:
cos(5x) < 0
To find where cos(5x) is negative, we look at the unit circle or the graph of cosine.
For cosine to be negative, the angle (5x) must be in the second and third quadrants (where cosine is negative). The second quadrant corresponds to an angle between π/2 and π, while the third quadrant corresponds to an angle between π and 3π/2.
So, for f(x) to be concave up, 5x must be in the interval (π/2, 3π/2).
Dividing the entire interval by 5, we get:
x ∈ (π/10, 3π/10)
Now let's find where f''(x) is negative (indicating concave down). We solve the inquality:
-25cos(5x) < 0
Dividing both sides by -25, we have:
cos(5x) > 0
For cosine to be positive, the angle (5x) must be in the first or fourth quadrants (where cosine is positive). The first quadrant corresponds to an angle between 0 and π/2, while the fourth quadrant corresponds to an angle between 3π/2 and 2π.
So, for f(x) to be concave down, 5x must be in the intervals [0, π/2) and (3π/2, 2π].
Dividing these intervals by 5, we have:
x ∈ [0, π/10) ∪ (3π/10, 2π/5]
Therefore, the open intervals where f(x) is concave up are (π/10, 3π/10) and the open intervals where f(x) is concave down are [0, π/10) ∪ (3π/10, 2π/5].