Question

Given f(x) 4 5 ​ x+2 a) Find f ′′ (x) b) Evaluate dx df ​ at x=3 and dx df −1 ​ at x=f(3) to show that at these points dx df −1 ​ =1/ dx df ​ . 2) Let f(x)=8x 3 −15x 2 −5,x≥1.5. Find the value of dx df −1 ​ at the point x=620=f(5).

Answer 1

`a) \(f''(x) = 48x - 30\)b) \( (dx)/(df) \Big|_(x=3) = (1)/(2) \), \( (dx)/(df^(-1)) \Big|_(x=f(3)) = 2 \)`

Step-by-step explanation:

`The second derivative of \(f(x) = 5x + 2\) is found by taking the derivative of the first derivative. Thus, \(f''(x) = 48x - 30\). To evaluate \( (dx)/(df) \) at \(x=3\), we substitute \(x=3\) into the derivative of \(f(x)\), giving \( (dx)/(df) \Big|_(x=3) = (1)/(2)\). For \( (dx)/(df^(-1)) \) at \(x=f(3)\), we find \(f(3) = 17\), and then evaluate the derivative of \(f^(-1)(x)\) at \(x=17\), yielding \( (dx)/(df^(-1)) \Big|_(x=f(3)) = 2\).`

`This shows that at these points, \( (dx)/(df^(-1)) = (1)/( (dx)/(df) )\), confirming the relationship.Moving on to the second part, for \(f(x) = 8x^3 - 15x^2 - 5\) with \(x \geq 1.5\), to find \( (dx)/(df^(-1)) \) at \(x=f(5)\), we first calculate \(f(5) = 620\). Then, evaluating the derivative of \(f^(-1)(x)\) at \(x=620\), we get \( (dx)/(df^(-1)) \Big|_(x=f(5)) = 2\).`