1. for this problem, consider the function ( f(x)=x^{4 / 3}+12 x^{1 / 3} ).\n(a) find the two critical…

1. for this problem, consider the function ( f(x)=x^{4 / 3}+12 x^{1 / 3} ).\n(a) find the two critical points of ( f(x) ).\n( f^{prime}=\frac{4}{3} x^{1 / 3}+\frac{12}{3} x^{-2 / 3} )\n( =\frac{4}{3} x^{1 / 3}+4 x^{-2 / 3} )\ncritical points:\n(b) use the first derivative test to classify each point you found in (a) as a local maximum, a local\nminimum, or neither.

1. for this problem, consider the function ( f(x)=x^{4 / 3}+12 x^{1 / 3} ).\n(a) find the two critical points of ( f(x) ).\n( f^{prime}=\frac{4}{3} x^{1 / 3}+\frac{12}{3} x^{-2 / 3} )\n( =\frac{4}{3} x^{1 / 3}+4 x^{-2 / 3} )\ncritical points:\n(b) use the first derivative test to classify each point you found in (a) as a local maximum, a local\nminimum, or neither.

Answer

Explanation:

Step1: Find the derivative of (F(x))

Using the power rule ((x^n)^\prime = nx^{n - 1}), for (F(x)=x^{4/3}+12x^{1/3}), we have (F^\prime(x)=\frac{4}{3}x^{\frac{4}{3}-1}+12\times\frac{1}{3}x^{\frac{1}{3}-1}=\frac{4}{3}x^{1/3}+4x^{- 2/3}=\frac{4x + 12}{3x^{2/3}})

Step2: Set (F^\prime(x) = 0) and find where (F^\prime(x)) is undefined

  • Set (F^\prime(x)=0): (\frac{4x + 12}{3x^{2/3}}=0). Since the denominator (3x^{2/3}\neq0) for (x\neq0), we solve (4x+12 = 0), which gives (x=-3)
  • Find where (F^\prime(x)) is undefined: The function (F^\prime(x)) is undefined when (x = 0) (because of the (x^{-2/3}=\frac{1}{x^{2/3}}) term). So the critical points are (x = 0) and (x=-3)

Step3: Use the First - Derivative Test

  • For (x=-3):
    • Choose a test point (x=-4) (left of (-3)): (F^\prime(-4)=\frac{4\times(-4)+12}{3\times(-4)^{2/3}}=\frac{-16 + 12}{3\times\sqrt[3]{16}}=\frac{-4}{3\times\sqrt[3]{16}}<0)
    • Choose a test point (x=-2) (right of (-3)): (F^\prime(-2)=\frac{4\times(-2)+12}{3\times(-2)^{2/3}}=\frac{-8 + 12}{3\times\sqrt[3]{4}}=\frac{4}{3\times\sqrt[3]{4}}>0) Since (F^\prime(x)) changes from negative to positive at (x=-3), (x = - 3) is a local minimum.
  • For (x = 0):
    • Choose a test point (x=-1) (left of (0)): (F^\prime(-1)=\frac{4\times(-1)+12}{3\times(-1)^{2/3}}=\frac{-4 + 12}{3\times1}=\frac{8}{3}>0)
    • Choose a test point (x = 1) (right of (0)): (F^\prime(1)=\frac{4\times1+12}{3\times1^{2/3}}=\frac{16}{3}>0) Since (F^\prime(x)) does not change sign at (x = 0), (x = 0) is neither a local maximum nor a local minimum

Answer:

(a) The critical points are (x = 0) and (x=-3). (b) (x=-3) is a local minimum and (x = 0) is neither.