For which values of \( x \) will the tangent to \( f(x)=-2 x^{3}-8 x \) have a positive gradient?

To find the values of \( x \) where the tangent to the function \( f(x) = -2x^3 - 8x \) has a positive gradient, we first need to determine the derivative \( f'(x) \). The derivative of the function is calculated as follows: \[ f'(x) = \frac{d}{dx}(-2x^3 - 8x) = -6x^2 - 8 \] Next, we set the derivative greater than zero to find where the gradient is positive: \[ -6x^2 - 8 > 0 \] To solve this inequality, we first rearrange it: \[ -6x^2 > 8 \quad \Rightarrow \quad x^2 < -\frac{8}{6} = -\frac{4}{3} \] However, since \( x^2 \) is always non-negative, there are no real values of \( x \) that satisfy this condition. Therefore, the derivative \( f'(x) \) is always negative, indicating that the tangent to the function does not have a positive gradient for any real value of \( x \). In conclusion, there are **no values of \( x \)** for which the tangent to the function \( f(x) = -2x^3 - 8x \) has a positive gradient.