Home > Calculus > Slope of a Curve at a Point

If #dy/dx = 6/(2x - 3)^2#, and #y(3) = 5#, what is the x-intercept of #y#?

Answer 1

Luke Alvarez

#(7/4, 0)#.

We will have to solve a differential equation to find the initial function. This d.e. is separable.

#dy/dx = 6/(2x -3)^2#

#dy = 6/(2x - 3)^2dx#

Integrate both sides.

#intdy = int 6/(2x - 3)^2 dx#

The right-hand side can be integrated by a u-substitution. Let #u = 2x - 3#. Then #du = 2dx# and #dx = (du)/2#.

#intdy =int 6/u^2 * (du)/2#

#intdy = int 3/u^2 du#

#intdy = 3u^-2#

#y = -3u^-1+ C#

#y = -3/u + C#

#y = -3/(2x - 3) + C#

Now solve for #C#. We know that when #x = 3#, #y = 5#.

#5 = -3/(2(3) - 3) + C#

#5 = -3/3 + C#

#5 = C - 1#

#C = 6#

Therefore, the initial function is #y = -3/(2x - 3) + 6#. The curve will cross the x-axis when #y = 0#.

#0 = -3/(2x - 3) + 6#

#-6 = -3/(2x - 3)#

#-6(2x - 3) = -3#

#-12x + 18 = -3#

#-12x = -21#

#x = 21/12#

#x = 7/4#

Hopefully this helps!

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.