f(x)=x^7(x+8)^8

Question

For -14<=x<=12 the function is defined by f(x)=x^7(x+8)^8 on which two intervals is the function increaseing , find the intervals which the function is positivie x= to x= where does the functino acieve it min x=

Answer 1

f(x) = x⁷(x + 8)⁸

We can take the derivative using the product rule (u•v)' = u'v + uv'

where:

u = x⁷

u' = 7x⁶

v = (x + 8)⁸

v' = 8(x + 8)⁷

So:

f '(x) = (7x⁶)(x + 8)⁸ + (x⁷)(8(x + 8)⁷

f '(x) = (7x⁶)(x + 8)⁸ + (8x⁷)(x + 8)⁷

f '(x) = (x⁶)(x + 8)⁷[7(x + 8) + 8x)]

f '(x) = (x⁶)(x + 8)⁷(15x + 56)

Setting this equal to zero, we can solve for the critical points by setting each factor equal to zero:

x⁶ = 0 when x = 0

(x + 8)⁷ = 0 when x = -8

15x + 56 = 0 when x = -56/15

Performing the first derivative test, we put these numbers on a number line and try numbers in each increment to see if the value of the derivative is positive or negative. Remember a positive 1st derivative means the function is increasing on that interval while a negative 1st derivative means the function is decreasing on that interval.

I arbitrarily picked values of -10, -5, -1, and 1 to be the values I would test:

f '(-10) = ((-10)⁶)(-10 + 8)⁷(15(-10) + 56) = +1.2x10¹⁰ - a positive so increasing on [-14, -8)

f '(-5) = ((-5)⁶)(-5 + 8)⁷(15(-5) + 56) = -6.5x10⁸ - a negative so decreasing on (-8, -56/15)

f '(-1) = ((-1)⁶)(-1 + 8)⁷(15(-1) + 56) = +3.4x10⁷ - a positive so increasing on (-56/15, 0)

f '(1) = ((1)⁶)(1 + 8)⁷(15(1) + 56) = +3.4x10⁸ - a positive so increasing on (0, 12]

This tell us that the function is increasing on [-14, -8) U (-56/15, 0) U (0, 12]

The 1st derivative tells us also how to categorize the critical points:

x = -8 is a local max

x = 56/15 is a local min

x = 0 is neither a min or a max

We are interested on the local minimum.

At x = -56/15 the function value is -1.11x10⁹

To verify if this is the minimum value, we also need to check the endpoints x = -14 and x = 12

f(-14) = -1.77x10¹⁴ which is a smaller value

f(12) is a large positive number because the function increases continually after x = -56/15

Therefore the minimum function value of -1.77x10¹⁴ occurs at x = -14

Answer 2

f(x) = x⁷(x+8)⁸ ffor x = [-14,12]

When is the function increasing? When f'(x)>0

f'(x) = 7x⁶(x+8)⁸ + 8x⁷(x+8)⁷

Which is 0 at x = 0 and -8

simplify by dividing by x⁶(x+8)⁶ (pos def) to obtain 7(x+8)² + 8x(x+8)

Clearly f'(x) >0 for all x >0 and x< -8, but solving the quadratic for other 0s, you get x = -56/15 as a 0 which means that we need to check around it. You find that x=-56/15 is a minimum and that the 0 point is an inflection point, so f'(x) > 0 for x = [-14,-8) U (-56/15,0) U (0,12)

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