Determine a simplified expression, for the slope of the secant #PQ# with #P(1, 1/3)# and #Q(1+h, f(1+h))# where #f(x) = 1/sqrt(2x+7)# where #x gt -7/2#. State restrictions on #h#?

1 Answer
Steve M
Feb 16, 2018

# "slope "PQ = (-2)/((3sqrt(2h+9))(3+sqrt(2h+9))) \ \ \ (h gt 0)#

Explanation:

We have;

# f(x) = 1/sqrt(2x+7) #

First we check that #P# lies on the given curve

# x=1=> f(1)= 1/sqrt(9) = 1/3 #

Confirming #P(1,1/3)# lies on the curve #y=f(x)#, along with #Q(1+h,f(1+h))#

Next we find the gradient of the chord #PQ#, which, providing #h gt 0#, is given by:

# m_(PQ) = (Delta y)/(Delta x) #

# \ \ \ \ \ \ \ = (f(1+h) - 1/3)/(1+h-1) #

# \ \ \ \ \ \ \ = (1/sqrt(2(1+h)+7) - 1/3)/(h) #

# \ \ \ \ \ \ \ = (3 - sqrt(2h+9))/(3hsqrt(2h+9)) * (3 + sqrt(2h+9))/(3 + sqrt(2h+9))#

# \ \ \ \ \ \ \ = (3^2 - (2h+9))/((3hsqrt(2h+9))(3+sqrt(2h+9))) #

# \ \ \ \ \ \ \ = (-2h)/((3hsqrt(2h+9))(3+sqrt(2h+9))) #

# \ \ \ \ \ \ \ = (-2)/((3sqrt(2h+9))(3+sqrt(2h+9))) #

Conclusion:
If we take the limit, as #h rarr 0#, then we get the slope of the tangent of the curve #y=f(x)# at #P#, (or the derivative ), which we can readily calculate:

# f'(1) = lim_(h rarr 0) m_(PQ) #

# \ \ \ \ \ \ \ \ = lim_(h rarr 0) (-2)/((3sqrt(2h+9))(3+sqrt(2h+9))) #

# \ \ \ \ \ \ \ \ = (-2)/((3sqrt(9))(3+sqrt(9))) #

# \ \ \ \ \ \ \ \ = (-2)/((9)(6)) #

# \ \ \ \ \ \ \ \ = -1/27 #

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