How do you find the inflection points of the graph of the function: #f(x)= (x^7/42) - (3x^6/10) + (6x^5/5) - (4x^4/3)#?

Question

Paisley Johnson · Accepted Answer

See the explanation.

For #f(x)= (x^7/42) - (3x^6/10) + (6x^5/5) - (4x^4/3)#, we get #f''(x) = x^5-9x^4+24x^3-16x^2# # = x^2( x^3-9x^2+24x-16)# Looking for rational zeros of the cubic, we note that #1# is a zro, so #x-1# is a factor. Dividing by #x-1# gets us: # = x^2(x-1) (x^2-8x+16)# # = x^2(x-1) (x-4)^2)# The zeros of #f''# are #0#, #1#, and #4#. The only one at which #f''# changes sign is #1#. The only inflection point is #(1,f(1))# (I'll leave it to the student to find #f(1)#)

Elijah Abram · Answer

undefined To find the inflection points of the function f(x) = (x^7/42) - (3x^6/10) + (6x^5/5) - (4x^4/3), follow these steps:

1. Compute the second derivative of the function f(x) with respect to x.
2. Set the second derivative equal to zero and solve for x to find the critical points.
3. Test each critical point by substituting it into the second derivative.
4. If the second derivative changes sign at a critical point, then it is an inflection point.

Let's proceed with the calculations:

1. Compute the second derivative of f(x):
   f'(x) = (7x^6/42) - (18x^5/10) + (30x^4/5) - (16x^3/3)
   f''(x) = (42x^5/42) - (90x^4/10) + (120x^3/5) - (48x^2/3)
          = x^5 - 9x^4 + 24x^3 - 16x^2

2. Set the second derivative equal to zero and solve for x:
   x^5 - 9x^4 + 24x^3 - 16x^2 = 0
   Factor out x^2: x^2(x^3 - 9x^2 + 24x - 16) = 0
   The solutions for x^2 = 0 are x = 0.

3. Now, we solve for the cubic equation x^3 - 9x^2 + 24x - 16 = 0.
   One of the solutions is x = 2.

4. Test each critical point:
   For x = 0: f''(0) = 0^5 - 9(0)^4 + 24(0)^3 - 16(0)^2 = 0. The test is inconclusive.
   For x = 2: f''(2) = 2^5 - 9(2)^4 + 24(2)^3 - 16(2)^2 = 32 - 9(16) + 24(8) - 16(4) = 32 - 144 + 192 - 64 = 16. The second derivative is positive.

Therefore, the inflection point is at x = 2.