How do you find the slope of the curve #f(x)=sqrt(x-1)# at the point x=5?

Question

Isabella Ackerman · Accepted Answer

The slope is #+- 1/4#
(there are two possible values as the function is not one-one as it contains a square root)

The slope of a function at some particular value of the independent variable is the first derivative evaluated at that particular value. #f(x)# is a compound function so it will be necessary to use the chain rule. Denoting the two component functions as #g(x)#, where #g(x) = sqrt(x)# and #h(x)#, where #h(x) = x - 1# it might be noted that #f(x) = g(h(x))# (#g# of #h# of #x#) The chain rule states #f'(x) = (g(h(x)))'# #= g'(h(x))h'(x) # That is, the derivative of the outer function evaluated at the inner function times the derivative of the inner function. It is convenient to write #g(x)# as #g(x) = (x)^(1/2)# so that the rules of polynomial differentiation might be easily applied #g'(x) = (1/2)(x)^(-1/2)# evaluating this at the value of the inner function #g'(h(x)) = (1/2)(x - 1)^(-1/2)# Noting #h'(x) = 1# The overall derivative is #f'(x) = (g(h(x)))' = g'(h(x))h'(x) = (1/2)(x - 1)^(-1/2)(1)# That is #f'(x) = 1/(2sqrt(x - 1))# so #f'(5) = 1/(2sqrt(5 - 1)) = 1/(2sqrt(4)) = +- 1/4#

Elijah Abram · Answer

undefined To find the slope of the curve $ f(x) = \sqrt{x-1} $ at the point $ x = 5 $, you can use the derivative of the function. The derivative of $ f(x) $ with respect to $ x $ is $ f'(x) = \frac{1}{2\sqrt{x-1}} $. Evaluating this derivative at $ x = 5 $ gives:

$$ f'(5) = \frac{1}{2\sqrt{5-1}} = \frac{1}{2\sqrt{4}} = \frac{1}{2 \times 2} = \frac{1}{4} $$

So, the slope of the curve $ f(x) = \sqrt{x-1} $ at the point $ x = 5 $ is $ \frac{1}{4} $.