Use the first derivative test to determine the location of each local extremum and the value of the function at this extremum. \( f(x)=x^{3} e^{4 x}+2 \)

To find the local extrema of the function \( f(x) = x^3 e^{4x} + 2 \) using the first derivative test, we first need to compute the derivative \( f'(x) \). After applying the product rule and chain rule, we get: \[ f'(x) = 3x^2 e^{4x} + 4x^3 e^{4x} \] Factoring out the common terms, we have: \[ f'(x) = e^{4x}(3x^2 + 4x^3) \] Setting the derivative equal to zero to find critical points: \[ e^{4x} \neq 0 \quad \Rightarrow \quad 3x^2 + 4x^3 = 0 \] Factoring this gives us: \[ x^2(3 + 4x) = 0 \] Thus, \( x = 0 \) and \( x = -\frac{3}{4} \) are critical points. To apply the first derivative test, we can analyze the sign of \( f'(x) \): 1. **For \( x < -\frac{3}{4} \)**: Choose \( x = -1 \): \( f'(-1) = e^{-4}(3 + 4(-1)) = e^{-4}(-1) < 0 \) (Decreasing) 2. **For \( -\frac{3}{4} < x < 0 \)**: Choose \( x = -0.5 \): \( f'(-0.5) = e^{-2}(3 + 4(-0.5)) = e^{-2}(1) > 0 \) (Increasing) 3. **For \( x > 0 \)**: Choose \( x = 1 \): \( f'(1) = e^{4}(3 + 4(1)) = e^{4}(7) > 0 \) (Increasing) From this analysis, we conclude: - Local minimum at \( x = -\frac{3}{4} \) since \( f'(x) \) changes from decreasing to increasing. - No local maximum as \( f'(x) \) is increasing for \( x > 0 \). Now we can plug the critical points back into \( f(x) \) to find the function values at these points. For \( x = -\frac{3}{4} \): \[ f\left(-\frac{3}{4}\right) = \left(-\frac{3}{4}\right)^3 e^{-3} + 2 = -\frac{27}{64} e^{-3} + 2 \] At \( x = 0 \): \[ f(0) = 0 + 2 = 2 \] Thus, the local extremum occurs at \( (-\frac{3}{4}, -\frac{27}{64} e^{-3} + 2) \) which is the local minimum, and at \( x = 0 \), the function value is 2.