Answer :

Answer:   34/49

Work Shown

[tex]f(x) = \frac{g(x)}{h(x)}\\\\f'(x) = \frac{g'(x)h(x)-h'(x)g(x)}{( h(x) )^2}\\\\f'(4) = \frac{g'(4)h(4)-h'(4)g(4)}{( h(4) )^2}\\\\f'(4) = \frac{1*3.5-(-1)*5}{( 3.5 )^2}\\\\[/tex]

[tex]f'(4) = \frac{8.5}{12.25}\\\\f'(4) = \frac{850}{1225}\\\\f'(4) = \frac{34*25}{49*25}\\\\f'(4) = \frac{34}{49}\\\\[/tex]

Notes:

  • I used the quotient rule on the 2nd step.
  • It's not entirely clear, but I'm assuming that h(4) = 3.5 since this y value appears to be at the midpoint of y = 3 and y = 4.
  • The derivative helps us find the slope of the tangent line. In the case of linear functions, the derivative is just the slope of the line itself.
  • 34/49 = 0.69387755 approximately