Is #f(x)=(x-3)(x+11)(x-7)# increasing or decreasing at #x=-1#?

Question

Christopher Agresta · Accepted Answer

f(x) is decreasing at x = -1 In this instance, it is probably "better" to distribute the brackets before differentiating than to apply the "product rule." (x - 3 )(x + 11 ) = # x^2+ 8x - 33# and #(x^2 + 8x -33 )(x - 7 )# # = x^3 - 7x^2 + 8x^2 - 56x - 33x + 231 # # = x^3 + x^2 - 89x + 231# You must look up the value of f'(-1) in order to determine whether the function is increasing or decreasing. • At x = -1, f(x) is increasing if f'(-1) > 0. • At x = -1, f(x) is decreasing if f'(-1) < 0. hence f(x) = #x^3 + x^2 - 89x + 231# so f'(x) # = 3x^2 + 2x -89# and f'(-1)# = 3(-1)^2 + 2(-1) - 89 = 3 - 2 - 89 = - 88 < 0 # f(x) is therefore decreasing at x = - 1.

Charles Anctil · Answer

undefined To determine if the function $ f(x) = (x - 3)(x + 11)(x - 7) $ is increasing or decreasing at $ x = -1 $, we need to evaluate the sign of the derivative of the function at that point. If the derivative is positive, the function is increasing at that point; if negative, the function is decreasing.

To find the derivative, we use the product rule:

$$ f'(x) = (x + 11)(x - 7) + (x - 3)(x - 7) + (x - 3)(x + 11) $$

Now, evaluate $ f'(-1) $:

$$ f'(-1) = (-1 + 11)(-1 - 7) + (-1 - 3)(-1 - 7) + (-1 - 3)(-1 + 11) $$
$$ = (10)(-8) + (-4)(-8) + (-4)(10) $$
$$ = -80 + 32 - 40 $$
$$ = -88 + 32 $$
$$ = -56 $$

Since $ f'(-1) = -56 $ is negative, the function $ f(x) $ is decreasing at $ x = -1 $.