Answer:
Show that if for some where , then by Rolle's Theorem for some . However, no such exists since for all .
Note that Rolle's Theorem alone does not give the exact value of the root. Neither does this theorem guarantee that a root exists in this interval.
Step-by-step explanation:
The function is continuous and differentiable over . By Rolle's Theorem. if for some where , then there would exist such that .
Assume by contradiction does have more than one roots over . Let and be (two of the) roots, such that . Notice that just as Rolle's Theorem requires. Thus- by Rolle's Theorem- there would exist such that .
However, no such could exist. Notice that , which is a parabola opening upwards. The only zeros of are and .
However, neither nor are included in the open interval . Additionally, , meaning that is a subset of the open interval . Thus, neither zero would be in the subset . In other words, there is no such that . Contradiction.
Hence, has at most one root over the interval .
Answer:
y = x - 19
Step-by-step explanation:
slope: 1/1x
(2,-17) Subtract by slope until x = 0, so you can get the y-intercept.
(1,-18)
(0,-19)
y - intercept: (0,-19)
Slope-intercept form: y = mx + b
y = x -19
Answer:
No answer.
Step-by-step explanation:
- 8m - 12 > -5m + 40 - 3m + 21
combine terms
-8m - 12 > -8m +61
add 8m to both sides
-12 > 61
negative 12 is not greater than 61