find all relative extrema of the function. use the second derivative test where applicable. (if an answer…

find all relative extrema of the function. use the second derivative test where applicable. (if an answer does not exist, enter dne.)\n\n$f(x)=x^{4}-8x^{3}+7$\n\nrelative maximum $(x,y)=(quad)$\n\nrelative minimum $(x,y)=(quad)$

find all relative extrema of the function. use the second derivative test where applicable. (if an answer does not exist, enter dne.)\n\n$f(x)=x^{4}-8x^{3}+7$\n\nrelative maximum $(x,y)=(quad)$\n\nrelative minimum $(x,y)=(quad)$

Answer

Explanation:

Step1: Find the first derivative

Using the power rule ((x^n)^\prime=nx^{n - 1}), for (y = f(x)=x^{4}-8x^{3}+7), we have (f^\prime(x)=4x^{3}-24x^{2}=4x^{2}(x - 6))

Step2: Find the critical points

Set (f^\prime(x)=0), so (4x^{2}(x - 6)=0). Solving (4x^{2}(x - 6)=0) gives (x = 0) or (x=6)

Step3: Find the second derivative

Differentiate (f^\prime(x)=4x^{3}-24x^{2}) using the power rule. (f^{\prime\prime}(x)=12x^{2}-48x=12x(x - 4))

Step4: Apply the second - derivative test

  • For (x = 0): Substitute (x = 0) into (f^{\prime\prime}(x)), (f^{\prime\prime}(0)=12\times0\times(0 - 4)=0). The second - derivative test is inconclusive. We can use the first - derivative test. For (x\lt0), say (x=-1), (f^\prime(-1)=4\times(-1)^{2}\times(-1 - 6)=4\times1\times(-7)=-28\lt0) For (0\lt x\lt6), say (x = 1), (f^\prime(1)=4\times1^{2}\times(1 - 6)=4\times1\times(-5)=-20\lt0) Since the sign of (f^\prime(x)) does not change around (x = 0), there is no relative extremum at (x = 0)
  • For (x = 6): Substitute (x = 6) into (f^{\prime\prime}(x)), (f^{\prime\prime}(6)=12\times6\times(6 - 4)=12\times6\times2 = 144\gt0) When (x = 6), (f(6)=6^{4}-8\times6^{3}+7=1296-1728 + 7=-425)

Answer:

relative maximum ((x,y)=\text{DNE}) relative minimum ((x,y)=(6,-425))