Question

Assume that \(f\) is differentiable for all \(x\). The signs of \(f^{\prime}\) are as follows. \(f^{\prime}(x)>0\) on \((-\infty,-4)\) \(f^{\prime}(x)<0\) on (-4,6) \(f^{\prime}(x)>0\) on \((6, \infty)\) Supply the appropriate inequality for the indicated value of \(c\). $$ g(x)=-f(x) \quad g^{\prime}(0) $$

Short answer

Therefore, using the information provided in the exercise and considering \(g(x)=-f(x)\), it can be concluded that \(g^{\prime}(0) > 0\).

Step-by-step solution

Understanding the Function

The function \(g(x)\) is given as \(g(x)=-f(x)\). So, the derivative of \(g(x)\) is going to be the opposite of the derivative of \(f(x)\), because the derivative of a constant multiplied function will be the constant multiplied by the derivative of the function.

Determine the Value of Derivative of Original Function

From the exercise, we know that \(f^{\prime}(x) < 0\) for the interval such that -4 < x < 6, and 0 is within that interval. Therefore, we can say that \(f^{\prime}(0) < 0\).

Calculate the Derivative of New Function

For the function \(g(x)\), since \(g(x)=-f(x)\), the derivative \(g^{\prime}(x)\) will be -\(f^{\prime}(x)\). So, \(g^{\prime}(0)\) will be -\(f^{\prime}(0)\). Since \(f^{\prime}(0) < 0\), the value of \(g^{\prime}(0)\) will be positive because -(-) is positive.