How many inflection points are in the graph of #f(x)= (x^7)/42 - (3x^6)/10 + (6x^5)/5 - (4x^4)/3#?

Question

Luke Alvarez · Accepted Answer

There is one inflection point.

#f(x)= (x^7)/42 - (3x^6)/10 + (6x^5)/5 - (4x^4)/3# #f'(x)= (x^6)/6 - (9x^5)/5 + 6x^4 - (16x^3)/3# #f''(x)= x^5 - 9x^4 + 24x^3 - 16x^2# Finding the zeros of #f''# #f''(x)= x^5 - 9x^4 + 24x^3 - 16x^2# # = x^2(x^3-9x^2+24x-16)# Rational Zeros Theorem and checking (or inspection of likely candidates or adding the coefficients) we see that #1# is a zero. Therefore #x-1# is a factor. Factor or do the division to get: #f''(x) = x^2(x-1)(x^2-8x+16)# # = x^2(x-1)(x-4)^2# The zeros are #0, 1, "and " 4#. The sign of #f''(x)# changes only at the single zero #x=1# Therefore the only inflection point is #(1, f(1))#.

Isabella Ackerman · Answer

undefined To find the inflection points, we need to determine where the concavity changes. This occurs when the second derivative changes sign. First, find the second derivative of f(x): f''(x) = (7x^5)/6 - (9x^4)/5 + (24x^3)/5 - (16x^2)/3. Now, set f''(x) equal to zero and solve for x to find potential inflection points. After solving, we find that there are 4 potential inflection points.