Question

The function f(x)=4+8 x+32 x⁻¹ has one local minimum and one local maximum. This function has a local maximum at x= ___ with value and a local minimum at x= ____ with value

Answer 1

The function f(x) = 4 + 8x + 32x⁻¹ has a local maximum at x = -1/2 with a value of -20 and a local minimum at x = 1/2 with a value of 68.

Step-by-step explanation:

The function in question is f(x) = 4 + 8x + 32x⁻¹. To find the local extremum, we calculate the first derivative f'(x) and set it to zero to find the critical points. Then, using the second derivative, we determine whether these critical points are local maxima or minima.

To find the derivative:
f'(x) = 8 - 32x⁻².
To find the critical points, set the derivative equal to zero:
8 - 32x⁻² = 0.
Solving for x gives x = ±rac{1}{2}.

Next, we check the second derivative, f''(x) = 64x⁻³.
At x = -1/2, f''(x) is negative, indicating a local maximum. At x = 1/2, it is positive, indicating a local minimum.

The local maximum is at x = -1/2 with value f(-1/2) = 4 + 8(-1/2) + 32(-2) = -20.
The local minimum is at x = 1/2 with value f(1/2) = 4 + 8(1/2) + 32(2) = 68.