The equation has no real roots, the minimum value is approximately f(x) = 1.2019577607847935 when
x = .6339
The equation can be simplified:
f(x) = x^7 (x-1) + 2x^5 (x-1) + 3x^3 (x-1) + 4x (x-1) + 5/2
f(x) = (x-1)*(x^7 + 2x^5 + 3x^3 + 4x) + 5/2
f(x) = x*(x-1)*(x^6 + 2x^4 + 3x^2 + 4) + 5/2
So g(x) = (f(x) - 5/2) has zeros at x=0 and x=1
So (0, 5/2) and (1, 5/2) are two points on the graph.
For x > 1, f(x) clearly only gets larger with larger x.
For x < 1, f(x) also clearly only gets larger with more negative x.
Take the derivative and set to zero to find the minimum:
f'(x) = 8x^7 - 7x^6 + 12*x^5 - 10*x^4 + 12*x^3 - 9*x^2 + 8*x - 4 = 0
Solved graphically, f'= 0 when x = 0.6339 and f(0.6339) = approx 1.2019577607847935
|
|
Posted by Larry
on 2020-11-06 07:36:18 |
Solution
