g(x)=x^3-5x^2-8 has x-intercept (there is only one) at appx. (5.2862792,0)
Use Calculus techniques to tell all you know about g(x).
Find the following:
a.) points where max and min occur
b.) intervals of concavity - concave up and concave down
c.) points of infllection
d.) Y- intercepts
e.) Sketch the graph and label important points (max, min, points of inflection)
Use Calculus techniques to tell all you know about g(x).
Find the following:
a.) points where max and min occur
b.) intervals of concavity - concave up and concave down
c.) points of infllection
d.) Y- intercepts
e.) Sketch the graph and label important points (max, min, points of inflection)
-
b)
g''(x) = 6x - 10
Where is g''(x) positive?
6x - 10 > 0
6x > 10
x > 10/6
x > 5/3
That's concave up.
Where is g''(x) negative?
6x - 10 < 0
6x < 10
x < 10/6
x < 5/3
That's concave down.
a)
g'(x) = 3x^2 - 10x
Set that to zero:
3x^2 - 10x = 0
x(3x - 10) = 0
x = 0 or x = 10/3
If x = 0, then g''(x) < 0, so the function is concave down, so that's a (local) maximum.
If x = 10/3, then g''(x) > 0, so the function is concave up, so that's a (local) minimum.
There is no global max/min.
c)
g''(x) = 6x - 10
Set that to zero:
6x - 10 = 0
6x = 10
x = 10/6
x = 5/3
d)
g(0) = -8
g''(x) = 6x - 10
Where is g''(x) positive?
6x - 10 > 0
6x > 10
x > 10/6
x > 5/3
That's concave up.
Where is g''(x) negative?
6x - 10 < 0
6x < 10
x < 10/6
x < 5/3
That's concave down.
a)
g'(x) = 3x^2 - 10x
Set that to zero:
3x^2 - 10x = 0
x(3x - 10) = 0
x = 0 or x = 10/3
If x = 0, then g''(x) < 0, so the function is concave down, so that's a (local) maximum.
If x = 10/3, then g''(x) > 0, so the function is concave up, so that's a (local) minimum.
There is no global max/min.
c)
g''(x) = 6x - 10
Set that to zero:
6x - 10 = 0
6x = 10
x = 10/6
x = 5/3
d)
g(0) = -8