Given #f(x) = x^2#, considering #f(4)#, #f(5)# and #f(6)#, how do you approximate the graph of #f(x)# by a straight line at #x=5# to deduce an approximation for #sqrt(25.3)# ?

1 Answer
George C.
Jan 8, 2018

#sqrt(25.3) ~~ 5.03#

Explanation:

Let:

#f(x) = x^2#

Note that:

#{ (f(4) = 16), (f(5) = 25), (f(6) = 36) :}#

If we differentiate #f(x)#, then we find:

#f'(x) = 2x#

So:

#f'(5) = 10#

Alternatively, if we approximate the slope at #(5, f(5))# by the slope of a secant through the points #(4, f(4)) = (4, 16)# and #(6, f(6)) = (6, 36)#, then we find:

#m = (Delta y)/(Delta x) = (color(blue)(36)-color(blue)(16))/(color(blue)(6)-color(blue)(4)) = 20/2 = 10#

By either method, we find a value #10# for the instantaneous slope at #5#. Note that if the curve was more complex than a parabola, then these values would normally differ.

To find the (principal) square root of #25.3# we want to find #x# such that #f(x) = 25.3#, i.e. #(x, 25.3)# is a point on our parabola.

Approximating the parabola near #(5, 25)# by a straight line of slope #10#, that gives us an approximation:

#sqrt(25.3) ~~ 5+(25.3-25)/10 = 5+0.3/10 = 5.03#

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