If some function f has the tangent line y − 2 = 4(t − 16) at the point implied by the equation, what are f (16) and f ′(16)?

f(16) = 2

f'(16) = 4

Step-by-step explanation:

Given:

- The function f(x) is given as:

                                   f(x) = y = 4t - 62

Find:

- what are f (16) and f ′(16)?

Solution:

- The value f(16) is given by plugging in x = 16 into the expression given as follows:

                                  f(16) = 4*16 - 62

                                  f(16) = 2

- The value of f'(16) is the gradient of the function @ x = 16.

The gradient is given by the derivative of the function:

                                  df(x) / dx = 4

Hence, f'(x) is in-dependent of the value of x and gradient is 4 for all values of x.

sorry i got nothing

step-by-step explanation:

2 over 5 and 4 over 15

don’t know last one sorry

step-by-step explanation: