How many points of inflection does the function #f(x) = (pi/3)^((x^3)-8)# have?

Question

Cameron Alexander · Accepted Answer

One. To find the points of inflection of any function, we compute the function's derivative then find the points where the derivative equals zero. In this case, we use the chain rule: let #u=x^3-8#, then #f'(x)=d/dx (pi/3)^(x^3-8)=(d/dx (pi/3)^u)(d/dx (x^3-8))#
#=>f'(x)=(pi/3)^(x^3-8) ln(pi/3)(3x^2)# There is no need to simplify this function any further. We know that for #f'(x)# to equal zero, then at least one of three things must happen: #(pi/3)^(x^3-8)=0#, or
#ln(pi/3)=0#, or
#3x^2=0#. Clearly, #ln(pi/3)# is not zero. Nor is #(pi/3)^(x^3-8)#, as any function #f(x)=a^(p(x))# is never zero for any polynomial #p(x)#. This leaves #3x^2#, and we know that #3x^2=0# only when #x=0#. Hence the one and only point of inflection is #x=0#.

Emily Ammerman · Answer

Two points of inflections:
#x=0 , f(0) = (pi/3)^(-8) " and " x=-2/ln(pi/3), f(-2/ln(pi/3)) = (pi/3)^(-8(1/(ln(pi/3))^3+1) # A point of inflection can be obtained when #f''(x) = 0#. Using the chain rule: let #u=x^3-8#, then #f(x)=(pi/3)^(x^3-8)# #f'(x)=d/dx (pi/3)^(x^3-8)=d/{du} (pi/3)^u {du}/{dx}# #=(pi/3)^(x^3-8) ln(pi/3)3x^2# #f''(x)=3ln(pi/3)(pi/3)^(x^3-8) x(x ln(pi/3) + 2) = 0# Since: #(pi/3)^(x^3-8)>0# only possible solution is when: #x(x ln(pi/3) + 2)=0#. Hence two points of inflections: #x=0 , f(0) = (pi/3)^(-8) " and " x=-2/ln(pi/3), f(-2/ln(pi/3)) = (pi/3)^(-8(1/(ln(pi/3))^3+1) # Note that when #x=0, f(0) # it is also a stationary point.

Christopher Agresta · Answer

undefined To find the points of inflection, we need to examine the second derivative of the function and locate where it changes sign. Taking the second derivative of $ f(x) = \left(\frac{\pi}{3}\right)^{x^3 - 8} $, we get:

$ f''(x) = \left(\frac{\pi}{3}\right)^{x^3 - 8} \cdot \ln^2\left(\frac{\pi}{3}\right) \cdot (x^3 - 8)^2 + 2\ln\left(\frac{\pi}{3}\right) \cdot \left(\frac{\pi}{3}\right)^{x^3 - 8} \cdot x^2(x^3 - 8) $

The second derivative can change sign when $ f''(x) = 0 $ or is undefined.

Setting $ f''(x) = 0 $ and solving for $ x $, we get:

$ \left(\frac{\pi}{3}\right)^{x^3 - 8} \cdot \ln^2\left(\frac{\pi}{3}\right) \cdot (x^3 - 8)^2 + 2\ln\left(\frac{\pi}{3}\right) \cdot \left(\frac{\pi}{3}\right)^{x^3 - 8} \cdot x^2(x^3 - 8) = 0 $

Since $ \left(\frac{\pi}{3}\right)^{x^3 - 8} $ is never zero, we can set the remaining part equal to zero:

$ \ln^2\left(\frac{\pi}{3}\right) \cdot (x^3 - 8)^2 + 2\ln\left(\frac{\pi}{3}\right) \cdot x^2(x^3 - 8) = 0 $

This equation has solutions for $ x $. After finding these solutions, we can test the sign of the second derivative around these points to determine if they are points of inflection.

However, since this involves complex calculations, it's better to use numerical methods or graphing software to approximate the points of inflection. Therefore, the number of points of inflection cannot be determined without further computation.