`f(x) = (x^3)(x - 2)^4` (a) Use a graph of `f` to give a rough estimate of the intervals of concavity and the coordinates of the points of inflection. (b) Use a graph of `f''` to give better estimates.
1 Answer | Add Yours
Using the graph of f(x), we determine the intervals of concavity based on the slopes of tangent line.
A decreasing slopes of tangent as we move to left to right (as x increase) indicates that the graph is concave down.
An increasing slopes of tangent as we move to left to right (as x increase) indicates that the graph is concave up
Slope of tangent line =0 at critical points (minimum, maximum or inflection point). Location of the critical point can be use as boundary values to set the intervals of the concavity.
This follows the first derivative test wherein:
Critical points occurs at x=a when f'(a) =0.
The graph for f(x) =` x^3(x-2)^4 ` is:
From the graph of f(x), the intervals of concavity are:
Concave up : (0, 0.8) and (2, +`oo` )
Concave down: (- `oo` , 0) and (0.8,2)
The critical points used as boundary values are:
inflection point at x=0
maximum point at x=0.8
minimum point at x=2
The graph of f"(x) = `6·x·(x - 2)^2·(7·x^2 - 12·x + 4)` or
`f(x) = 42x^5 - 240x^4 + 480x^3 - 384x^2 + 96x`
Using the graph of the second derivative to find the intervals of concavity, we follow:
f"(x) > 0 or graph of f"(x) below the x-axis for concave up
f"(x)< 0 graph of f"(x) above the x-axis for concave down
f"(x) =0 possible on inflection point.
Real inflection point occurs at x=a when there is change in concavity before and after x=a.
As shown in the graph, possible inflection points occurs at
x-values: 0, 0.45, 1.25, and 2 where the graph f"(x) intersects the x-axis.
Intervals of concavity:
Concave up: (0,0.45), (1.25 , +`oo` )
Concave down:(-`oo` , 0), (0.45 , 1.25).
Based on this, there are a real inflection points at x=0, 0.45, 1.26.
We’ve answered 318,926 questions. We can answer yours, too.Ask a question