# f(x)=xsqrt(x+4)A. find the interval of increase and decrease. b. find the local minimum values c. find the interval where the graph is concave upwards.

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### 1 Answer

`f(x)= xsqrt(x+4) `

`==gt f'(x)= sqrt(x+4) + x/(2sqrt(x+4))`

` ==gt f'(x)= (2x+8 + x)/(2sqrt(x+4) `

`==gt f'(x)= (3x+8)/(2sqrt(x+4)) `

`==gt (3x+8) = 0 ==gt x = -8/3 `

`==gt f(-8/3) = (-8/3)sqrt(-8/3 + 4) `

`= (-8/3)sqrt(-8+12)/3`

`= -8/3*2/sqrt3 `

`= -16/3sqrt3`

= -3.08

**Then, the minimum value is f(-8/3)= -3.08**

**==> f(x) is decreasing for the interval (-4, -8/3) **

**==> f(x) is increasing for the interval (-8/3, `oo` ).**