im having so much trouble figuring out how to get these answers. i already know the answers i just need to figure out how to get them...
directions: find the points of inflection and discuss the concavity of the graph of the function...
1. f(x)=(1/2)x^4+2x^3
2. f(x)=x^3-6x^2+12x
3. f(x)=(1/4)x^4-2x^2
4. f(x)=2sinx+sin2x, [0,2ð)
1.
f = 1/2 x^4 + 2x^3
f' = 2x^3 + 6x^2
f'' = 6x^2 + 12x
f' = 0 at x=0,0,-3 so the possible mins/maxes are there
f'' = 0 at 0,-2 so the inflection is there.
At x=0, f'=0 and f''=0, so not a min/max -- inflection
At x=-3, y'' > 0, so concave up (min)
Proceed similarly with the others.
rechneronline . de is a good grapher
Finding the points of inflection and discussing the concavity of a graph requires some knowledge of calculus. You'll need to use the second derivative of the function to determine the concavity and find the points of inflection.
Here's how you can approach each of the given functions:
1. To find the points of inflection and discuss the concavity of the graph of f(x) = (1/2)x^4 + 2x^3, follow these steps:
a. Find the first derivative of the function: f'(x) = 2x^3 + 6x^2.
b. Find the second derivative of the function by differentiating f'(x): f''(x) = 6x^2 + 12x.
c. Set the second derivative equal to zero and solve for x to find any potential points of inflection: 6x^2 + 12x = 0. The solutions are x = 0 and x = -2.
d. To discuss the concavity, analyze the sign of the second derivative. For x < -2, f''(x) > 0, which means the graph is concave up. For -2 < x < 0, f''(x) < 0, so the graph is concave down. And for x > 0, f''(x) > 0, indicating the graph is concave up again.
2. To find the points of inflection and discuss the concavity of the graph of f(x) = x^3 - 6x^2 + 12x, follow these steps:
a. Find the first derivative of the function: f'(x) = 3x^2 - 12x + 12.
b. Find the second derivative of the function by differentiating f'(x): f''(x) = 6x - 12.
c. Set the second derivative equal to zero and solve for x to find any potential points of inflection: 6x - 12 = 0. The solution is x = 2.
d. To discuss the concavity, analyze the sign of the second derivative. For x < 2, f''(x) < 0, indicating the graph is concave down. For x > 2, f''(x) > 0, so the graph is concave up.
3. To find the points of inflection and discuss the concavity of the graph of f(x) = (1/4)x^4 - 2x^2, follow these steps:
a. Find the first derivative of the function: f'(x) = x^3 - 4x.
b. Find the second derivative of the function by differentiating f'(x): f''(x) = 3x^2 - 4.
c. Set the second derivative equal to zero and solve for x to find any potential points of inflection: 3x^2 - 4 = 0. The solutions are x = -2/√3 and x = 2/√3 (approximately ±1.154).
d. To discuss the concavity, analyze the sign of the second derivative. For x < -2/√3 and x > 2/√3, f''(x) > 0, which means the graph is concave up.
4. To find the points of inflection and discuss the concavity of the graph of f(x) = 2sin(x) + sin(2x), within the interval [0, 2π] (assuming ð represents π), follow these steps:
a. Find the first derivative of the function: f'(x) = 2cos(x) + 2cos(2x).
b. Find the second derivative of the function by differentiating f'(x): f''(x) = -2sin(x) - 4sin(2x).
c. Set the second derivative equal to zero and solve for x to find any potential points of inflection: -2sin(x) - 4sin(2x) = 0. There can be multiple solutions within the interval [0, 2π].
d. To discuss the concavity, analyze the sign of the second derivative. If f''(x) > 0, the graph is considered concave up, and if f''(x) < 0, it is concave down.
Remember to sketch the graph of each function to better visualize concavity and points of inflection.