Is #f(x)=(x+1)(x+5)(x-7)# increasing or decreasing at #x=-1#?

Question

Charles Arocha · Accepted Answer

Decreasing

While you can figure out the answer by calculating the derivative #f^prime (-1)# the traditional way, the following approach (which works when you are trying to find the derivative of a function at one of its zeroes) is often quicker. When #x# is close to -1, we have #x+5 ~~ -1+5 = 4# and #x-7 ~~ -1-7 = -8#. So, in the immediate vicinity of #-1#, the function is approximately #f(x) ~~ (x+1) times 4 times (-8) = -32(x+1)# As a result, the function is getting smaller. In fact, this method also gives you the value of #f^prime(-1)# - it is easy to see that the value is -32.

Emily Ammerman · Answer

undefined To determine whether $ f(x) = (x+1)(x+5)(x-7) $ is increasing or decreasing at $ x = -1 $, we need to evaluate the sign of the derivative of $ f(x) $ at that point. If the derivative is positive, $ f(x) $ is increasing; if it's negative, $ f(x) $ is decreasing.

To find the derivative of $ f(x) $, we can use the product rule and the chain rule. After finding the derivative, we substitute $ x = -1 $ into the derivative expression to determine its sign.

After computing the derivative of $ f(x) $, we find $ f'(x) = (x+5)(x-7) + (x+1)(x-7) + (x+1)(x+5) $.

Now, substituting $ x = -1 $ into $ f'(x) $, we get $ f'(-1) = (-1+5)(-1-7) + (-1+1)(-1-7) + (-1+1)(-1+5) $.

Evaluating this expression gives us $ f'(-1) = (4)(-8) + (0)(-8) + (0)(4) = -32 $.

Since $ f'(-1) = -32 $ is negative, $ f(x) $ is decreasing at $ x = -1 $.