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Calculus - Minimum Value of a Function










































Geometrically, the 1st derivative of a function is the slope of the curve.

When the 1st derivative is positive, the function is increasing.

When the 1st derivative is negative, the function is decreasing.

When the 1st derivative is zero, the function has reached a maximum, minimum, or an inflection point.

When the 1st derivative is zero. the 2nd derivative will show whether that point is a maximum, minimum, or an inflection point.

The 2nd derivative is the slope of the 1st derivative function.

The 2nd derivative is negative when the original function has reached a maximum (at that point where the 1st derivative is zero).

The 2nd derivative is positive when the original function has reached a minimum (at that point where the 1st derivative is zero).

The 2nd derivative is zero when the original function has reached a point of inflection (at that point where the 1st derivative is zero).


Determine the minimum value of the function y=3x^3+x^2-15x+2 for x<0.

Comments for Calculus - Minimum Value of a Function

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Mar 06, 2013
Minimum Value
by: Staff


Answer


Part I


a plot of the function f(x) = 3x³ + x² - 15x + 2 is shown below

Graph of the third degree function f(x) = 3x³ + x² - 15x + 2





There is no minimum value when x < 0.

The minimum value occurs when x > 0.

However, there is a maximum value when x < 0. The maximum value occurs when x ≈ -1.4068781988545.

1st & 2nd derivatives of y = 3x³ + x² - 15x + 2

First and Second Derivatives of f(x) = 3x³ + x² - 15x + 2




the value of x when the 1st derivative is zero

First Derivative Test of f(x) = 3x³ + x² - 15x + 2:  relative maxima and minima





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Mar 06, 2013
Minimum Value
by: Staff


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Part II

2nd derivative of f(x) = 3x³ + x² - 15x + 2

y'' = 18x + 2

when x = x1

Second Derivative Test applied to critical point x_1 of f(x) = 3x³ + x² - 15x + 2






Because the 2nd derivative is a negative value when x ≈ -1.4068781988545, this point is a maximum


when x = x2

Second Derivative Test applied to critical point x_2 of f(x) = 3x³ + x² - 15x + 2




Because the 2nd derivative is a positive value when x ≈ 1.1846559766323, this point is a minimum





Thanks for writing.

Staff
www.solving-math-problems.com



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