Consider the function f(x)= (x^7)(e^(3x)), -3 < or equal to x > or equal to 41) The absolute minimum value is 2) and this occurs at x equals

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to determine the critical values of the function over interval [-3,4], hence you need to solve the equation f'(x) = 0 such that:

`f'(x) = 7x^6*e^(3x) + 3x^7*e^(3x)`

You need to solve the equation f'(x) = 0 such that:

`7x^6*e^(3x) + 3x^7*e^(3x) = 0`

You need to factor out `x^6*e^(3x)`   such that:

`x^6*e^(3x) (7 + 3x)=0 `

Since e^(3x)!=0,then you should solve `x^6 = 0`  and `7 + 3x = 0`  such that:

`x^6 = 0 =gt x = 0`

`7 + 3x = 0 =gt 3x = -7 =gt x = -7/3`

You need to select a value for x in interval `[-3,0]`  such that:

`x = -1 =gt f'(-1) = (-1)^(6)*(1/e^3)(7 - 3) = 4/e^3 gt 0`

You need to select a value for x in interval `[-3,-7/3] ` such that:

`x = -3 =gt (-3)^6*(1/e^9)* (7 - 9) lt 0`

You need to select a value for x in interval `[0,4]`  such that:

`x = 1 =gt 1^6*e^(3) (7 + 3)gt0 `

Hence, the function reaches its absolute minimum at `x= -7/3` .

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