1 Answer | Add Yours
`f(x) = 2x^4-11x^3+17x^2`
To find the extreme points. you should first find the first derivative and then analyze thier nature using the second derivative.
`f'(x) = 8x^3-33x^2+34x`
For extreme points, `f'(x) = 0,`
`8x^3-33x^2+34x = 0`
`x(8x^2-33x+34) = 0`
This gives, `x = 0` or `8x^2-33x+34 = 0.`
`8x^2-33x+34 = 0`
`x = (33+-sqrt(33^2-4xx8xx34))/(2xx8)`
`x = (33+-sqrt(1))/16`
This gives, x = 2 or x = 2.125.
Therefore the values for extreme points are, x = 0, x=2 and x=2.125.
To assess their nature we need to find the second derivative.
`f''(x) = 24x^2-66x+34`
At x = 0, f''(0) > 0. Therefore at x = 0 f(x) has minimum with,
`f(0) = 0-0+0 = 0`
Therefore (0,0) is a local minimum of f(x).
At x =2, f''(2) <0. Therefore at x =2 f(x) has a maximum with,
`f(2) = 2xx2^4-11xx2^3+17xx2^2`
`f(2) = 12.`
Therefore, (2,12) is local maximum of f(x)
At x = 2.125, f''(2.125) > 0. Therefore at x=2.125 f(x) has a minimum with,
`f(2.125) = 2xx(2.125)^4-11xx(2.125)^3+17xx(2.125)^2`
`f(2.125) = 11.99463`
Therefore, (2.125,11.99463) is a local minimum of f(x).
From `-oo` to `0` -
f(x) is concave up and decreasing
From `0` to `2`
f(x) is concave up and increasing
From `2` to `2.125`
f(x) is concave down and decreasing
From `2.125` to `+oo`
f(x) is concave up and increasing.
Note that around x=2, it is not an inflection point, you just need to zoom extremely to see the maxiimum and minimum. Try MATLAB, you will see them.
We’ve answered 318,005 questions. We can answer yours, too.Ask a question