Question

The graph of function f is shown above. if g(x)=x^2/f(x) what is the value of g'(1)?​

Answer 1

Answer:

Step-by-step explanation:

Find the equation of the segment going from (0,-5) to (3,7)

y intercept = -5

Slope = (-5 - 7) / (0 - 3) = -12/-3 = 4

equation: y = 4x - 5

g(x) = x^2 / f(x)

f(x)=  (4x - 5)

g(x) = x^2 / (4x - 5)

g'(x) = x^2 * (4x - 5)^-1

g'(x) = 2x*(4x - 5)^-1  + (-1) *4* x^2 (4x - 5)^-2

I will leave that monster the way it is and just find g'(1)

g'(1) = 2(1) * (4(1) - 5)^-1    +    (-1) (1)^2 *4* (4(1) - 5)^-2

g'(1) = 2(1) * (-1)^-1             +    (-1) (1)^2 *4 * (-1)^2

g'(1) = -2                      +        (-1) (1)^2 (4)

g'(1) = - 2                     +        (-1) (1)^2 (4)

g'(1) =  - 2 - 4

g'(1) = - 6