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

We are given   f(x) = x³e^(4x) + 2. Step 1. Find the first derivative. Differentiate f(x) using the product rule on x³e^(4x):   d/dx [x³e^(4x)] = (x³)'e^(4x) + x³(e^(4x))'               = 3x²e^(4x) + x³·4e^(4x)               = e^(4x)[3x² + 4x³]. So,   f'(x) = e^(4x)[3x² + 4x³]        = e^(4x)x²(3 + 4x). Since the constant 2 differentiates to 0, we only have the term from x³e^(4x). Step 2. Determine the critical points. A critical point occurs when f'(x) = 0 or f'(x) is undefined. Here, e^(4x) is never zero, so f'(x) = 0 when:   x²(3 + 4x) = 0. This gives: 1. x² = 0 → x = 0 (a repeated root), 2. 3 + 4x = 0 → x = -3/4. Thus, the candidates for local extrema are x = -3/4 and x = 0. Step 3. Apply the First Derivative Test. We need to examine the sign of f'(x) in intervals determined by these critical points. Note: f'(x) = e^(4x)x²(3 + 4x). Since e^(4x) > 0 and x² ≥ 0 for all x, the sign of f'(x) is determined solely by the factor (3 + 4x). • For x < -3/4:   Choose x = -1.   3 + 4(-1) = 3 - 4 = -1. (Negative)   Thus f'(x) is negative for x < -3/4. • For -3/4 < x < 0:   Choose x = -0.5.   3 + 4(-0.5) = 3 - 2 = 1. (Positive)   Thus f'(x) is positive for -3/4 < x < 0. • For x > 0:   Choose x = 1.   3 + 4(1) = 3 + 4 = 7. (Positive)   Thus f'(x) is positive for x > 0. Analysis of the sign changes: 1. At x = -3/4:   f'(x) changes from negative to positive. Therefore, by the first derivative test, x = -3/4 is a local minimum. 2. At x = 0:   f'(x) is positive just to the left and positive to the right (since x² causes a zero derivative at x = 0 but does not change the sign of (3+4x)). Hence, there is no change in the monotonicity of f(x), and x = 0 is not a local extremum (in fact, it is a stationary inflection point). Step 4. Find the function value at the local minimum. At x = -3/4, compute f(-3/4):   f(-3/4) = (-3/4)³ e^(4(-3/4)) + 2        = (-27/64)e^(-3) + 2. Thus, the local minimum occurs at x = -3/4 with   f(-3/4) = 2 - (27e^(-3))/64. Final Answer: The function f(x) = x³e^(4x) + 2 has a local minimum at x = -3/4, and its value at this point is f(-3/4) = 2 - (27e^(-3))/64. There is no local maximum.