Sketch the graph and show all local extrema and inflection points.
f(x)= 1/x^2-2x-8
I believe the ans is "Relative max: [1,-1/9]" and No inflection points".
f(x) = x^-2 - 2x - 8
f ' (x) = -2x^-3 - 2
f '' (x) = 6x^-4
extrema:
-2/x^3 = 2
x^3 = -1
x = -1, f(-1) = 1 - 2(-1) - 8 = -5
at x = -1, f ''(-1) is > 0, so
(-1,-5) is a minimum
my answer confirmed by
http://www.wolframalpha.com/input/?i=plot++y+%3D+1%2Fx%5E2-2x-8
pts of inflection:
6/x^4 = 0
so solution, thus no points of inflection.
Good solution, but wrong problem.
f = 1/(x^2-2x-8) = 1/((x-4)(x+2))
f' = 2(x-1)/(x^2-2x-8)^2
f" = 6(x^2-2x+4)/(x^2-2x-6)^3
f'=0 at x=1
f" is never zero, so no inflection points.
We see there are vertical asymptotes at x=4 and x=-2
f'=0 at x=1
f"(1) < 0, so f(1) is a local max
take it to wolframalpha to see the graph.
Oh yeah. Forgot to mention that y=0 is the horizontal asymptote.
To sketch the graph of the function f(x) = 1/x^2 - 2x - 8 and determine the local extrema and inflection points, you can follow these steps:
Step 1: Find the critical points by taking the derivative of the function. The critical points occur where the derivative equals zero or is undefined. So, let's differentiate f(x) first:
f'(x) = 0
Using the Quotient Rule, we can differentiate 1/x^2:
f'(x) = (-2/x^3) - 2
Set f'(x) equal to zero and solve for x:
(-2/x^3) - 2 = 0
-2/x^3 = 2
-2 = 2x^3
x^3 = -1
x = -1
So, we have one critical point at x = -1.
Step 2: Determine the nature of the critical point by finding the second derivative. Differentiate f'(x) to get f''(x):
f''(x) = 6/x^4
Since f''(x) is defined for all x ≠ 0, we don't have any inflection points since they occur when the second derivative equals zero.
Step 3: Determine the local extrema by evaluating f(x) at the critical point and the endpoints of the domain if they exist. Since f(x) is defined for all x ≠ 0, we only need to evaluate f(x) at x = -1.
f(-1) = 1/(-1)^2 - 2(-1) - 8
f(-1) = 1 - (-2) - 8
f(-1) = 1 + 2 - 8
f(-1) = -5
So, we have a relative maximum at (-1, -5).
Step 4: Sketch the graph.
The graph will approach positive and negative infinity as x approaches positive and negative infinity, respectively. However, since there are no asymptotes mentioned, we won't add them to the graph.
The graph passes through the point (-1, -5), which is a relative maximum.
Based on the analysis, there are no inflection points in the graph.
Therefore, the final answer is that the graph of f(x) = 1/x^2 - 2x - 8 has one relative maximum at (-1, -5) and no inflection points.