concavity and second derivative

Question

on the interval [-6,6] f(x) is continuous and differentiable. If f'(x)= (x²-4)(x+1) briefly justify the following conclusion:

 

"f'(x) has an x intercept at x=-1 but x=-1 is not a relative extrema on the graph of f(x) because..."

 

Not sure how to answer this problem.. could someone please show me the steps too? I think I will understand better if I see how it is solved :) Also, are they wanting it in sentence form?

Answer 1

We can use the first and second derivative test.  First derivative test is when we set f'(x) equal to zero.  Second derivative test is when we set f''(x) equal to zero.

 

First derivative test:

 

(x² - 4)(x + 1) = 0

 

(x + 2)(x - 2)(x + 1) = 0

 

x = -2  ,   x = -1  ,    x = 2

 

These x values are your critical points and location of local maximum and local minimums.

 

 

We will first expand f'(x), so that we can easily derive f'(x) to get f''(x).

 

f'(x) = x³ + x² - 4x - 4

 

 

f''(x) = 0

 

3x² + 2x - 4 = 0

 

 

Use the quadratic formula to solve for x.

 

x = (-2 ± √(4 - 4(-12))) / 6

 

x = (-2 ± √(4 + 48)) / 6

 

x = (-2 ± √(52)) / 6

 

x = -1.535        and         x = 0.869

 

 

These are your points of inflection.  They are the location where f(x) changes concavity.  Next, we use test points to evaluate the second derivative.  If the second derivative is negative, then it is concave down and indicates a maximum.  If the second derivative is positive, then it is concave up and indicates a minimum.

 

Evaluate  f''(-2)  ,  f''(0)  ,   f''(1)

 

f''(-2) = 3(-2)² + 2(-2) - 4

         = 12 - 4 - 4

        = 4

 

f''(0) = -4

 

f''(1) = 3(1)² + 2(1) - 4

        = 3 + 2 - 4

        = 1

 

Intervals of concave up are (-∞, -1.535)∪(0.869, ∞)

Interval of concave down are (-1.535, 0.869) 

 

We know that between -1.535 and 0.869, there is a maximum.  x=-1 lies in this interval.  However this is not enough to determine if x=-1 is a local extrema.  We go back to the first derivative and use test points.  If the first derivative changes signs, then we have a local extrema.

 

Evaluate  f'(-1.5) and  f'(0)  since x=-1  is between  x=-1.5  and  x=0

 

f'(-1.5) = (-1.5 + 2)(-1.5 - 2)(-1.5 + 1)

           = (0.5)(-2.5)(-0.5)

           = positive number

 

f'(0) = (0 + 2)(0 - 2)(0 + 1)

       = (2)(-2)(1)

       = negative number

 

Since the first derivative changes signs, x=-1 is a relative extrema.

 

 

I think you wrote your statement in the question incorrectly.

Answer 2

Emma,

I must admit that I am not sure what is wanted here. But I believe the point is that f(-1) is neither the highest nor the lowest on the interval [-6,6].  The easiest way to prove this to yourself is to calculate the anti-derivative and plug in all the critical points as well as the end points of the interval.  Of course we have no means to find C, but that's okay because it doesn't affect the relationships between the parts of the function.

 

f'(x) = x³ + x² - 4x -4

 

f(x) = x⁴/4 + x³/3 - 2x² - 4x + C

 

ignoring C we get f(-6) = 204, f(-2) = 4/3,  f(-1) = 23/12, f(2) = -28/3,  f(6) = 300

 

So for the interval in question, f(2) is the relative low and f(6) is the relative high. So a possible answer is:

.....because f(-1) is neither the highest nor the lowest point on the interval [-6,6}.