Answer :
Answer:
The second derivative test states that if the second derivative is positive at the point of a local extrema, it is a minimum; if it is negative, the point is a maximum; and if it equals zero, the test is inconclusive. Using this test, we find that this function has a local maximum at x=0 and a local minimum at x=32/9.
Step-by-step explanation:
Step one - Find the local extrema:
Find the local extrema by using the first derivative. We find the derivative of x⁹- 4x⁸ using the power rule, which states that the derivative of xⁿ = nxⁿ⁻¹. So, the derivative of this function is 9x⁸-32x⁷.
The local extrema of this function are located at the zeros of the derivative. Factor to find the zeros:
[tex]9x^8-32x^7=x^7(9x-32) = 0\\x=0,\frac{32}{9}[/tex]
So, the local extrema of this function are found at 0 and 32/9.
Step two - Classify the extrema:
To classify the extrema, we will use the second derivative. The second derivative test states that if the second derivative is positive at the point of the extrema, it is a minimum; if it is negative, the point is a maximum; and if it equals zero, the test is inconclusive.
The second derivative of this function is 72x⁷-224x⁶. We will test x=0, 32/9
f''(0) = 0, so the test is inconclusive and we must use the first derivative test. The first derivative test tests values surrounding the point to determine if it is a min or max. For this example, I will test -1 and 1.
f'(-1) = 9+32 = 41 -- since it is positive, the function is increasing here.
f'(1) = 9-32= -23 -- since it is negative, the function is decreasing here.
Because the function goes from increasing to decreasing, x=0 is a maximum.
f''(32/9) ≈6,500 -- since it is positive, x=32/9 is a minimum.
