Given the domain of the function `f(x)=x^2*sqrt(x+2)` is (-1,2) find the maximum and minimum values.

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The function `f(x) = x^2*sqrt(x + 2)` is defined for values of x lying in (-1,2). The maximum and minimum values of the function has to be determined.

To find the extreme points of the function `f(x) = x^2*sqrt(x + 2)` determine the derivative f'(x) and solve f'(x) = 0 for x.

f'(x) = `2x*sqrt(x + 2) + x^2*(1/2)*(1/sqrt(x + 2))`

=> `(2x*sqrt(x+2)*2*sqrt(x+2) + x^2)/(2*sqrt(x+2))`

=> `(4x*(x+2)+x^2)/(2*sqrt*x+2))`

=> `(5x^2 + 8x)/(2*sqrt*x+2))`

f'(x) = 0

=> `(5x^2 + 8x)/(2*sqrt*x+2)) = 0`

=> `5x^2 + 8x = 0`

This gives x = 0 and x = `-8/5` of which x = 0 lies in the domain. At x = 0, f(x) = 0.

The maximum value of x is obtained at the extreme of the domain. At x = 2. f(x) = `4*sqrt 4` = 8. But x = 2 does not lie in the domain. The maximum value of the function is a value infinitesimally smaller than 8.

The minimum value of f(x) is 0 and the maximum value is one that is infinitesimally smaller than 8.

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sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

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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.

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