Concavity and second derivative

Question

I am having issues with the following two problems:

 

Find the critical number of f, describe the open intervals on which f is increasing or decreasing and locate all relative extrema of

f(x)=x²+8x+10

f(x)= (x+3)/(x² )

Answer 1

When we set the first derivative of a function equal to zero, we determine the original function's critical points.

Let's take a look at the first equation:

 

f(x) = x² + 8x + 10

 

The first derivative would be:

 

f'(x) = 2x + 8

 

Now if we set this equal to zero and solve for x:

 

2x + 8 = 0

2x = -8

x = -4

 

So the critical point occurs at x = -4. Now, since we know from the original formula that f(x) is an upward-opening parabola (the coefficient in front of the x is positive), we can conclude that the critical point at x = -4 is the vertex of this concave-up parabola.

 

However, if we were unable to deduce this from the equation itself, we could determine it by means of the second-derivative test. First we would need to take the derivative again, this time of the first derivative.

 

f'(x) = 2x + 8

f''(x) = 2

 

Now, because f'(x) was a linear function, we will get the second derivative as a number. In this case it is positive so that tells us that the original function f(x) is concave up.

 

To locate the relative extrema, we simply plug -4 into the original function:

 

f(x) = x² + 8x + 10

f(-4) = (-4)² + 8(-4) + 10

f(-4) = 16 - 32 + 10

f(-4) = -6

 

So the vertex is located at (-4, -6)

 

Not all critical points are vertices though. In some cases they are inflection points, or where the function changes its concavity. In that case, you would need to select a point on either side of each of the critical points and test them with the second derivative test (Note: if the function changes concavity, its second derivative will contain the variable x). If the second derivative test yields a positive number, the concavity is up. If it yields a negative number, the concavity is down.

Answer 2

Critical numbers are typically peaks, valleys, and inflection points. Peaks and valleys are found by setting the first derivative equal to zero. The inflection points are found by setting the second derivative equal to zero.

 

f(x) = x² + 8x + 10

 

f'(x) = 2x + 8                    f"(x) = 2

 

2x + 8 = 0                         f"(x) never equals zero, no inflection points

2x = -8

x = -4

 critical point is (-4, -6)

 

test points around critical point for negative or positive slope:

 

f'(-5) = -2     f'(-3) = 2

 

decreasing (-∞,-4)     increasing (-4,∞)

 

absolute minimum at x = -4

 

_____________________________________________

 

f(x) = (x+3)/x²

 

For this one, you may also want to note the horizontal asymptote at y=0, and the vertical asymptote at x=0, it depends on the teacher.

 

f'(x) = [x² - 2x(x + 3)]/x⁴ = (-x² - 6x)/x⁴ = (-x - 6)/x³

 

setting this equal to zero requires only the numerator:  -x - 6 = 0  ⇒  x = -6

 

f"(x) = [-x³ - 3x²(-x - 6)]/x⁶ = (2x³ + 18x²)/x⁶ = (2x + 18)/x⁴

 

again setting the numerator equal to zero:  2x + 18 = 0  ⇒  x = -9

 

decreasing (-∞,-6), (0,∞)       increasing (-6,0)

 

inflection point at x = -9