What does the 2nd Derivative Test tell you about the behavior of #f(x) = x^4(x-1)^3# at these critical numbers?

Question

Christopher Agresta · Accepted Answer

The Second Derivative Test implies that the critical number (point) #x=4/7# gives a local minimum for #f# while saying nothing about the nature of #f# at the critical numbers (points) #x=0,1#.

If #f(x)=x^4(x-1)^3#, then the Product Rule says #f'(x)=4x^3(x-1)^3+x^4*3(x-1)^2# #=x^3*(x-1)^2*(4(x-1)+3x)# #=x^3*(x-1)^2*(7x-4)# Setting this equal to zero and solving for #x# implies that #f# has critical numbers (points) at #x=0,4/7,1#. Using the Product Rule again gives: #f''(x)=d/dx(x^3*(x-1)^2) * (7x-4)+x^3*(x-1)^2*7# #=(3x^2*(x-1)^2+x^3*2(x-1)) * (7x-4) + 7x^3 * (x-1)^2# #=x^2 * (x-1) * ((3x-3+2x) * (7x-4) + 7x^2-7x)# #=x^2 * (x-1) * (42x^2-48x+12)# #=6x^2 * (x-1) * (7x^2-8x+2)# Now #f''(0)=0#, #f''(1)=0#, and #f''(4/7)=576/2401>0#. The Second Derivative Test therefore implies that the critical number (point) #x=4/7# gives a local minimum for #f# while saying nothing about the nature of #f# at the critical numbers (points) #x=0,1#. In actuality, the critical number (point) at #x=0# gives a local maximum for #f# (and the First Derivative Test is strong enough to imply this, even though the Second Derivative Test gave no information) and the critical number (point) at #x=1# gives neither a local max nor min for #f#, but a (one-dimensional) "saddle point".

Charles Arocha · Answer

undefined The Second Derivative Test tells us about the behavior of a function at critical points by analyzing the concavity of the function. For the function $ f(x) = x^4(x-1)^3 $, we first find the critical points by setting the first derivative equal to zero and solving for $ x $. Then, we evaluate the second derivative at these critical points.

The critical points for $ f(x) $ occur where $ f'(x) = 0 $. So, we find $ f'(x) $ by taking the derivative of $ f(x) $ and setting it equal to zero:
$$ f'(x) = 4x^3(x-1)^3 + x^4(3)(x-1)^2 = 0 $$

After solving for $ x $, we obtain the critical points. Then, we evaluate the second derivative $ f''(x) $ at each critical point.

The second derivative of $ f(x) $ is:
$$ f''(x) = 12x^2(x-1)^3 + 12x^2(x-1)^2 + 4x^3(3)(x-1)^2 + 2x^4(3)(x-1) $$

After evaluating $ f''(x) $ at each critical point, we can determine the behavior of $ f(x) $ at those points:
- If $ f''(x) > 0 $ at a critical point, then the function is concave up at that point, indicating a local minimum.
- If $ f''(x) < 0 $ at a critical point, then the function is concave down at that point, indicating a local maximum.
- If $ f''(x) = 0 $ at a critical point, the test is inconclusive.

This information helps us understand the behavior of $ f(x) $ around its critical points and whether those points correspond to local maxima or minima.