calculus Question

Question

Use the second derivative test to find the maximum and minimum points of f(x) = 2x³+15x²-36x.

Help!?

Answer 1

To solve this problem, let us first take the first and second derivatives of f(x).

f'(x)=6x²+30x-36. f''(x)=12x+30.

To find where maxima/minima occur on a continuous, differential function, we must find where the slope/ first derivative =0. We can find them with simple factoring.

6x2+30x-36=6(x²+5x-6)=6(x+6)(x-1). So the slope =0 when x=-6 or 1.

Let's start with -6. If we plug it into f''(x), we get a negative number. This shows that the graph of f is concave downward at that point, showing that there is a maximum at x=-6. Plugging in 1 to f''(x) gives a positive number, so it is a minimum. Plus -6 and 1 into f(x) if you want to find y at those points. Hope this helped!

Answer 2

f'(x) = 6x² + 30x - 36

 

f"(x) = 12x + 30

 

 

Set the first derivative equal to zero.

 

6(x² + 5x - 6) = 0

3(x + 6)(x - 1) = 0

 

x = -6           and           x = 1

 

 

Set the second derivative equal to zero.

 

12x + 30 = 0

6(2x + 5) = 0

 

x = -5/2 = -2.5

 

The function is concave down in the interval (-∞, -2.5) and concave up in the interval (-2.5, ∞).  Based on this,

 

Maximum at f(-6).

Minimum at f(1).