Suppose that it is given to you that
f′(x)=(x+3)(8−x)(x−15)
The first inflection point (from the left) for f(x) occurs at x=
The second inflection point (from the left) for f(x) occurs at x=
inflection points occur where f"(x) = 0
f"(x) = -(3x^2-40x+51)
So, just find the roots of that to get the inflection points for f(x)
Suppose that f(x)=x^[1/3](x+3)^[2/3]
(A) Find all critical values of f. If there are no critical values, enter None . If there are more than one, enter them separated by commas.
Critical value(s) =
(B) Use interval notation to indicate where f(x) is increasing.
(C) Use interval notation to indicate where f(x) is decreasing.
Decreasing:
(D) Find the x-coordinates of all local maxima and minima of f
(E) Use interval notation to indicate where f(x) is concave up and concave down.
(F) Find all inflection points of f.
Post a new question please
To find the inflection points of a function, you need to determine where the concavity changes. This can be done by analyzing the second derivative of the function.
Given that f'(x)=(x+3)(8−x)(x−15), we can find the second derivative f''(x) by differentiating f'(x) with respect to x.
First, let's find the derivative of (x+3)(8−x)(x−15) using the product rule:
f''(x) = [(x+3)(-1) + (8-x)(1)][(x−15)] + (x+3)(8−x)[1]
Simplifying this expression, we get:
f''(x) = (9-2x)(x−15) + (x+3)(8−x)
Expanding and collecting like terms, we have:
f''(x) = 9x − 135 − 2x^2 + 30x + 8 − x^2 + 3x
Combining like terms again, we get:
f''(x) = -3x^2 + 42x − 127
Now that we have the second derivative, we can find the inflection points by solving the equation f''(x) = 0.
-3x^2 + 42x − 127 = 0
This equation can be solved using factoring, completing the square, or the quadratic formula. In this case, let's solve it using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In the equation -3x^2 + 42x − 127 = 0, a = -3, b = 42, and c = -127.
Plugging these values into the quadratic formula, we get:
x = (-(42) ± √((42)^2 - 4(-3)(-127))) / (2(-3))
Simplifying further:
x = (-42 ± √(1764 - 1524)) / (-6)
x = (-42 ± √240) / (-6)
x = (-42 ± √(16 * 15)) / (-6)
x = (-42 ± 4√15) / (-6)
Finally, we have the two possible inflection points:
x = (-42 + 4√15) / (-6)
x ≈ -10.08
and
x = (-42 - 4√15) / (-6)
x ≈ 17.42
Therefore, the first inflection point (from the left) for f(x) occurs at x ≈ -10.08, and the second inflection point (from the left) for f(x) occurs at x ≈ 17.42.