concavity, max, min, and inflection points

Question

Answer the following questions for the function

 

f(x) = xsqrt(x^2+4)

defined on the interval [-6,7]

 

A. f(x) is concave down on the region ____ to______
B. f(x) is concave up on the region  ____ to______
C. The inflection point for this function is at
D. The minimum for this function occurs at
E. The maximum for this function occurs a

Answer 1

Hi Adam,

 

Let us begin your problem by finding the second derivative of the given function.  The first step is to find f'(x) using the product rule.  This is given by

 

f'(x) = (1) · (x² + 4)^1/2 + (x) · (2x) · (1/2) · (x² + 4)^-1/2

    

       = (x² + 4)^1/2 + x · (x² + 4)^-1/2

 

At this point, notice that this function is continuous and always positive (this fact will be useful later).  Next, we take another derivative giving us f ''(x).

 

f ''(x) = (1/2) · (2x) · (x² + 4)^-1/2 + (2x) · (x² + 4)^-1/2 - (x²) · (2x) · (1/2) · (x² + 4)^-3/4

 

            =  (3x) · (x² + 4)^-1/2 - (x³) · (x² + 4)^-3/4

 

We can simplify this one step further by finding a common denominator.  This will result in the following

 

f ''(x) = (2x³ + 12x) · (x² + 4)^-3/4

 

Now is a good point to notice that the denominator of this expression is always positive (so we don't have any divide by zero situations) and the numerator is just a degree 3 polynomial, therefore the function is continuous.  Next we notice that the numerator is a degree three polynomial that can be factored as 

 

(2x)(x² + 6)

 

If we set this equal to zero (to find the points of inflection), then we get only one solution.  This solution comes from solving 2x = 0.  This solution is simply 

 

x = 0

 

Why no other solutions? Well if we try to solve x² + 6 = 0, we get imaginary solutions (try it with quadratic formula).  Now we can consider the inflection point theorem that says:

 

If f'(x) exists and f ''(x) changes sign at a point x=c, then the point (c , f(c)) is an inflection point.  Further, if f ''(x) also exists, it must be equal to zero.

 

Lets take inventory of what we have done learned about this function so far

 

1. f'(x) is continuous so it exists everywhere.
2. f ''(x) is continuous so it exists everywhere.
3. f ''(0) = 0

 

Now if we choose test point x = -1 and x = 1.  We have

 

f ''(-1) = -14 · (5^-3/2)< 0

 

and

 

f ''(1) = 14 · (5^-3/2)> 0

 

Since f ''(x) changes sign at x=0, we can conclude that this corresponds to an inflection point.  Therefore our function is

1. concave down from -∞ to 0,
2. concave up from 0 to ∞,
3. it has an inflection point at (0,0),
4. and it does not have a maximum or minimum because it is always decreasing (since the derivative is always positive).

 

 

 

Answer 2

A and B.

According to Concavity Theorem: If a function f(x) is twice differentiable, then the graph of f(x) is concave upward at f''(x) > 0 and concave downward at f''(x) <0

 

So, we need to calculate the second derivative of f(x)

 

f'(x) = d/dx (x√(x^2+4)) = 2(x^2 + 2)/√(x^2 + 4)

f''(x) = d/dx {2(x^2 + 2)/√(x^2 + 4)} = 2x(x^2 + 6)/(x^2 + 4)^(3/2)

 

Now, f''(x)>0 when x > 0, because (x^2 + 6) and (x^2 + 4)^(3/2) are always positive

Similarly, f''(x)<0 when x < 0

 

Since the domain of function f(x) is restricted to [-6, 7], f(x) would be concave up in the region from 0 to 7 and concave down in the region from -6 to 0

 

C. Inflection point is when f''(x) = 0

2x(x^2 + 6)/(x^2 + 4)^(3/2) = 0

x = 0

 

D and E.

Since f(x) is concave down in the region [-6, 0) and concave up in the region (0, 7], the maximum value would occur  at x = 7 and minimum would occur at x = -6

 

 

Answer 3

To find max and min, we need the first derivative.

 

f(x) = xsqrt(x²+4) = x(x²+4)^1/2

 

f ' (x) = x*1/2(x²+4)^-1/2 * 2x + (x²+4)^1/2 * 1 = x²/sqrt(x²+4) + sqrt(x^2+4)

 

= x²/sqrt(x²+4) + (x²+4)/sqrt(x²+4) = (2x²+4)/sqrt(x²+4)

 

Then we set f ' (x) = 0 (or where f ' (x) is undefined and f(x) is defined) and solve for x to find the critical points:

 

(2x²+4)/sqrt(x²+4) = 0

 

This will never = 0 (or be undefined) so there are no critical points.  Then all we have to do is plug our interval endpoints into the original function and see where the max and min are.

 

f(-6) = (-6)sqrt((-6)²+4) = (-6)sqrt(36+4) = -6sqrt40 = -12sqrt10 = MINIMUM

f(7) = (7)sqrt((7)²+4) = (7)sqrt(49+4) = 7sqrt53 = MAXIMUM

 

Now for concavity and inflection points, we need to find f '' (x).

 

f ' (x) = (2x²+4)/sqrt(x²+4)

 

f '' (x) = [sqrt(x²+4)*(4x)-(2x²+4)*1/2(x²+4)^-1/2(2x)]/(x²+4) = [(x²+4)(4x)-(2x²+4)x]/(x²+4)^3/2

 

= (4x³+16x-2x³-4x)/(x²+4)^3/2 = (2x³+12x)/(x²+4)^3/2

 

Now set f '' (x) = 0 (or if f '' (x) is undefined where f(x) is defined) to find the inflection points:

 

(2x³+12x)/(x²+4)^3/2 = 0

 

The only way this can equal 0 is if the numerator = 0.  f '' (x) can't be undefined as the denominator can't = 0.

 

2x³+12x=0

2x(x²+6)=0

 

2x=0

x = 0

 

So we have an inflection point at x = 0.

 

Now, for concavity, we check the sign of f '' (x) around the inflection point x = 0.

 

So to the left of x = 0, let's try x = -1

 

f '' (-1): numerator = 2(-1)³+12(-1) = -2-12 = -14 = negative; denominator = ((-1)²+4)^3/2 = (1+4)^3/2 = positive

 

f '' (-1) = negative/positive = negative

 

Since f '' (x) is negative to the left of our inflection point, that means it is concave down from (-6,0)

 

To the right of x = 0, let's try x = 1

 

f '' (1): numerator = 2(1)³+12(1) = 2+12 = 14 = positive; denominator = ((1)²+4)^3/2 = (1+4)^3/2 = positive

 

f '' (1) = positive/positive = positive

 

Since f '' (x) is positive to the right of our inflection point, that means it is concave up from (0,7)