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设函数f ( x ) = 2x + \frac { a } { x ^ { 2 } } ( a > 0 )在( 0 , + \infty )内的最小值为f ( ...

May 13, 2024

设函数f ( x ) = 2x + \frac { a } { x ^ { 2 } } ( a > 0 )在( 0 , + \infty )内的最小值为f ( x _ { 0 } ) =\newline 12,则x _ { 0 } =\newline

Solution by Steps

step 1

To find the minimum of the function

f(x) = 2x + \frac{a}{x^2}

for x > 0, we use calculus methods

step 2

We take the derivative of

f(x)

with respect to

x

to find the critical points:

f'(x) = 2 - \frac{2a}{x^3}

step 3

Set the derivative equal to zero to find the critical points:

2 - \frac{2a}{x^3} = 0

step 4

Solve for

x

to find the critical points:

x = \sqrt[3]{\frac{a}{1}}

step 5

Substitute the critical point back into

f(x)

to find the minimum value:

f(\sqrt[3]{a}) = 2\sqrt[3]{a} + \frac{a}{(\sqrt[3]{a})^2} = 3\sqrt[3]{a}

step 6

Given that the minimum value of

f(x)

is 12, we set

3\sqrt[3]{a} = 12

and solve for

a

step 7

Solving for

a

gives

a = 64

step 8

Now, we solve for

x_0

using the equation

12 = 2x_0 + \frac{64}{x_0^2}

step 9

Rearrange the equation to form a cubic equation in terms of

x_0

x_0^3 - 6x_0^2 + \frac{64}{2} = 0

step 10

Solve the cubic equation to find the value of

x_0

[1] Answer

x_0 = 4

Key Concept

Optimization using calculus and solving cubic equations

Explanation

To find the minimum of a function, we take its derivative, set it to zero, and solve for the critical points. Then we check which of these points gives the minimum value of the function. In this case, we also used the given minimum value to find a specific parameter

a

, and then solved a cubic equation to find

x_0

Generate me a similar question

How to find the derivative of the function

g(x) = 3x^2 + \frac{b}{x}

for x > 0?