  # 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

 Mar 06, 2013 Minimum Value by: Staff Answer Part I a plot of the function f(x) = 3x³ + x² - 15x + 2 is shown below 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 the value of x when the 1st derivative is zero ----------------------------------------------------------

 Mar 06, 2013 Minimum Value by: Staff ---------------------------------------------------------- Part II 2nd derivative of f(x) = 3x³ + x² - 15x + 2 y'' = 18x + 2 when x = x1 Because the 2nd derivative is a negative value when x ≈ -1.4068781988545, this point is a maximum when x = x2 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