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
