Find all critical points and the use the first-derivative test to determine local maxima and minima. 1) \( f(x)=3 x^{4}-4 x^{3}+6 \) 2) \( f(x)=\left(x^{3}-8\right)^{4} \)

Alright, I need to find all critical points of the given functions and determine whether each critical point is a local maximum, local minimum, or neither using the first-derivative test. Let's tackle each function one by one. **First Function: \( f(x) = 3x^4 - 4x^3 + 6 \)** 1. **Find the first derivative \( f'(x) \):** \[ f'(x) = \frac{d}{dx}(3x^4) - \frac{d}{dx}(4x^3) + \frac{d}{dx}(6) = 12x^3 - 12x^2 + 0 = 12x^3 - 12x^2 \] 2. **Set the first derivative equal to zero to find critical points:** \[ 12x^3 - 12x^2 = 0 \\ 12x^2(x - 1) = 0 \] So, the critical points are at \( x = 0 \) and \( x = 1 \). 3. **Determine the nature of each critical point using the first-derivative test:** - **At \( x = 0 \):** - Choose a test point to the left of 0, say \( x = -1 \): \[ f'(-1) = 12(-1)^3 - 12(-1)^2 = -12 - 12 = -24 < 0 \] - Choose a test point to the right of 0, say \( x = 1 \): \[ f'(1) = 12(1)^3 - 12(1)^2 = 12 - 12 = 0 \] Since the derivative changes from negative to zero, \( x = 0 \) is a **local minimum**. - **At \( x = 1 \):** - Choose a test point to the left of 1, say \( x = 0 \): \[ f'(0) = 12(0)^3 - 12(0)^2 = 0 \] - Choose a test point to the right of 1, say \( x = 2 \): \[ f'(2) = 12(2)^3 - 12(2)^2 = 96 - 48 = 48 > 0 \] Since the derivative changes from zero to positive, \( x = 1 \) is a **local minimum**. **Second Function: \( f(x) = (x^3 - 8)^4 \)** 1. **Find the first derivative \( f'(x) \):** \[ f'(x) = 4(x^3 - 8)^3 \cdot \frac{d}{dx}(x^3 - 8) = 4(x^3 - 8)^3 \cdot 3x^2 = 12x^2(x^3 - 8)^3 \] 2. **Set the first derivative equal to zero to find critical points:** \[ 12x^2(x^3 - 8)^3 = 0 \] This equation is zero when either \( x^2 = 0 \) or \( (x^3 - 8)^3 = 0 \). - \( x^2 = 0 \) implies \( x = 0 \). - \( (x^3 - 8)^3 = 0 \) implies \( x^3 - 8 = 0 \), so \( x = 2 \). Therefore, the critical points are at \( x = 0 \) and \( x = 2 \). 3. **Determine the nature of each critical point using the first-derivative test:** - **At \( x = 0 \):** - Choose a test point to the left of 0, say \( x = -1 \): \[ f'(-1) = 12(-1)^2((-1)^3 - 8)^3 = 12(1)(-9)^3 = 12 \cdot (-729) = -8748 < 0 \] - Choose a test point to the right of 0, say \( x = 1 \): \[ f'(1) = 12(1)^2((1)^3 - 8)^3 = 12(1)(-7)^3 = 12 \cdot (-343) = -4116 < 0 \] Since the derivative does not change sign around \( x = 0 \), \( x = 0 \) is neither a local maximum nor a local minimum. - **At \( x = 2 \):** - Choose a test point to the left of 2, say \( x = 1 \): \[ f'(1) = 12(1)^2((1)^3 - 8)^3 = 12(1)(-7)^3 = 12 \cdot (-343) = -4116 < 0 \] - Choose a test point to the right of 2, say \( x = 3 \): \[ f'(3) = 12(3)^2((3)^3 - 8)^3 = 12(9)(19)^3 = 108 \cdot 6859 = 740,412 > 0 \] Since the derivative changes from negative to positive, \( x = 2 \) is a **local minimum**. **Summary:** - For \( f(x) = 3x^4 - 4x^3 + 6 \): - **Critical Points:** \( x = 0 \) and \( x = 1 \) - **Nature:** Both are local minima. - For \( f(x) = (x^3 - 8)^4 \): - **Critical Points:** \( x = 0 \) and \( x = 2 \) - **Nature:** \( x = 0 \) is neither, and \( x = 2 \) is a local minimum.