Question

i used (f(x+h)-f(x))/h method but got a really messy answer

estim ts) Find the equation of the tangent line to the graph of the function f(x) = 73 – 5x at the point (,1). scratchwork frath) - flasin W3tx flath) - a $13

Answer 1

fin= √3-5x, Polet (²1) f'(22) = - Sk = Slope question ;-) = -52 21-45)

Answer 2

SOLUTION :

f(x) = sqrt(3 - 5x) = (3 - 5x)^(1/2)

So, f’(x) 

= (1/2) (3 - 5x)^(1/2 - 1) * d/dx (3 - 5x)

= 1/(2 sqrt(3 - 5x)) * (- 5) 

Multiply numerator and denominator by sqrt(3 - 5x) :

= - 5/2 * sqrt(3 - 5x) / (3 - 5x) 

Slope of tangent at point (2/5, 1) = f’(x) at (2/5, 1)

=> slope of tangent at (x = 2/5) = - 5/2 * sqrt(3 - 5 * 2/5) / (3 - 5*2/5) = - 5/2 *1 / 1 = - 5/2 

So, equation of tangent will be,

y = - 5/2 x + c 

As point(2/5, 1) is on tangent also, it should satisfy the above equation.

So,

1 = - 5/2 * 2/5 + c = - 1 + c

=> c = 2 

Hence, equation of tangent will be :

y = - 5/2 x + 2  (ANSWER).