find the intervals of increase and decrease,all local extrema,the intervals of concavity,all inflection points and sketch graph f(x)=2x^4-11x^3+17x^2
`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.