You need to solve the equation `f'(x) = 0` to find the extreme values of function, hence you should find first derivative f'(x) using product law of derivatives such that:
`f'(x) = 2xsqrt(x+2) + x^2/(2sqrt(x+2))`
You need to solve the equation f'(x) = 0 such that:
`2xsqrt(x+2) + x^2/(2sqrt(x+2)) = 0`
`(4x(x+2) + x^2)/(2sqrt(x+2)) = 0`
Since `(2sqrt(x+2)) != 0` , then `4x(x+2) + x^2 = 0` such that:
`4x^2 + 8x+ x^2 = 0`
Collecting like terms yields:
`5x^2 + 8x = 0`
You need to factor out x such that:
`x(5x + 8) = 0`
`x_1 = 0`
`5x + 8 = 0 =gt x_2 = -8/5 = -1.6`
Since -1.6 `!in` `(-1,2),` then the only extreme value of function is at `x=0` .
You need to select a value smaller than 0, x = -1/2, and you need to evaluate `f'(-1/2) = (5/4 - 4)/(2sqrt(-1/2+2)) lt 0` .
You need to select a value larger than 0, x = 1,and you need to evaluate `f'(1) = (5+8)/(2sqrt3) gt 0` .
Hence, the function decreases over (-1,0) and the function increases over (0,2), thus, the function reaches its minimum at x=0 and its maximum at x=2.