What  is the minimum value of the expression f(x) = x^3 - 5x + 8

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william1941 | College Teacher | (Level 3) Valedictorian

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Now we have to find the minimum value of f(x) = x^3 - 5x +8.

For that we start with finding the derivative of f(x) = x^3 - 5x +8

f’(x) = [x^3 + 5x +8]’ = 3x^2 + 5

Now equate this to 0

=> 3x^2 - 5 = 0

=> x^2 = 5/3

=> x = -sqrt [5/3] or +sqrt [5/3]

Now take the second derivative of f(x)

f’’(x) = 6x

for x= -sqrt [5/3], 6x = -6*sqrt [5/3] which is negative, therefore the maximum value is at x= -sqrt [5/3]

for x= +sqrt [5/3] , 6x = +6*sqrt [5/3] which is positive, therefore the minimum value is at x = +sqrt [5/3]

The minimum value is: f[(+sqrt [5/3])] = (+sqrt [5/3])^3 – 5*(+sqrt [5/3]) +8 = 3.69

The required result is 3.69

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

f(x) = x^3-5x+8. To find the minimum.

We know that f(x) is minimum  at a point x = c, for which f'(c) = 0 and f"(c) < 0.

f'(x) = (x^3-5x+8)' = 3x^2-5.

f'(x) = 0 gives 3x^2-5 = 0. c = sqrt5/3 Or  c = -sqrt5.

f"(x) = (3x^2-5)' = 6x.

f"( (sqrt(5/3)) = 6*sqrt(5/3) > 0.

Therefore f(sqrt(5/3))  = (sqrt(5/3))^3 - 5sqrt(5/3)+8 is minimum

Therefore min f(x) = f (sqrt(5/3)) = (5/3) sqrt5 - 5sqrt(5/3)+8 = 8 - (10/3) sqrt(5/3) = 8-(10/9)sqrt15 = 3.6967.

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