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2 Answers
Ratnaker Mehta
May 7, 2018

#(a) : (3,27/2); (b) : y=27/2, or, 2y-27=0; (c) : (1,55/2)#.

Explanation:

Part (a) :

At the stationary points, #dy/dx=0#.

#y=(x^3+54)/(2x)=x^3/(2x)+54/(2x)=x^2/2+27/x#.

#:. dy/dx=(2x)/2+27(-1/x^2)=x-27/x^2.............(diamond)#.

#:. dy/dx=0 rArr x-27/x^2=0 rArr x^3-27=0#.

#:. x=3" is the only real root."#

#x=3, y=(x^3+54)/(2x)=(3^3+54)/(2xx3)=27/2#.

Hence, #(3,27/2)# is a stationary point.

Part (b) :

Since at the stationary point, #dy/dx=0#. Hence, the slope of the

tangent at that point is zero. So, it is a horizontal line

through the stationary point #(3,27/2)#.

Thus, the eqn. the tgt. is #y=27/2, or, 2y-27=0#.

Part (c) :

Suppose that, at the point #(x_0,y_0), x_0 gt 0#, the gradient to the

curve is #-26#.

But, the gradient of the curve at #(x_0,y_0)# is #[dy/dx]_{P(x_0,y_0)}#.

#:. x_0-27/x_0^2=-26............[because, (diamond)]#.

#:. (x_0^3-27)=-26x_0^2, or, x_0^3+26x_0^2-27=0#.

Observe that, the sum of the co-effs. is #0#.

#:. (x_0-1)" is a factor"#.

#"Now, "x_0^3+26x_0^2-27=0 rArr(x_0-1)(x_0^2+27x_0+27)=0#

#:. x_0=1, x_(01,02)=(-27+-sqrt(27^2-4*1*27))/2#.

#:. x_0=1, x_(01)=(-27+3sqrt69)/2 lt 0, x_02=(-27-3sqrt69)/2 lt 0#.

Since reqd. #x gt 0, x_01 and x_02# are inadmissible.

The corresponding #y_0=(x_0^3+54)/(2x_0)=(1+54)/2=55/2.#

Hence, #(1,55/2)# is the desired point.

Andrea S.
May 7, 2018

a) #{(barx=3),(bary=27/2):}#

b) #y=27/2#

c) #x=1#

Explanation:

A stationary point is a point where the derivative of the function vanishes. Given then the equation of the curve:

#y = (x^3+54)/(2x)#

evaluate its first derivative using the quotient rule:

#dy/dx = (2x * 3x^2 - 2(x^3+54))/(4x^2)#

#dy/dx = (6x^3 - 2x^3-108)/(4x^2)#

#dy/dx = (x^3 -27)/x^2#

and solve the equation:

#dy/dx = 0#

#(x^3 -27)/x^2 = 0#

#x^3 -27 = 0 # with #x != 0#

#x=3#

The corresponding value of #y# is:

#y(3) = (3^3+54)/(2*3) = (27+54)/6 = 27/2#

The coordinates of the stationary point #P(barx, bary)# are then;

#{(barx=3),(bary=27/2):}#

By definition in a stationary point the tangent is horizontal because the derivative is null, so the equation of the tangent to the curve in #P# is:

#y=27/2#

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Finally, the value of the gradient of a scalar function is the value of its derivative so we can find the point where the value of the gradient is #-26# by solving the equation:

#dy/dx = -26#

#(x^3 -27)/x^2 = -26#

#x^3 +26x^2 -27 = 0#

We can easily see that #x=1# is a solution and dividing by #(x-1)# we have:

#(x^2+27x+27)(x-1) = 0#

and as the roots of:

#x^2+27x+27= 0#

must be negative, we can conclude that the solution is #x=1#.

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