Answer:
(a) p = 54
(b) k = 12
Step-by-step explanation:
Question 22(a)
The given curve is defined by the function
[tex]y = x^2- \dfrac{p}{x}[/tex]
The turning point of a curve defined by f(x) (if it exists) is located at the minimum or maximum value of that function.
The corresponding coordinates of the turning point can be computed by
(a) finding the first derivative of f(x)
(b) setting the first derivative = 0 and solving for x
(c) setting the value of x obtained in step (b) into the original equation and solving
Given
[tex]y = x^2- \dfrac{p}{x}\\\\f'(x) = \dfrac{d}{dx}\left(x^2 - \dfrac{p}{x}\right)\\\\= \dfrac{d}{dx}\left(x^2 \right) - \dfrac{d}{dx}\left(\dfrac{p}{x}\right)\\\\[/tex]
[tex]\dfrac{d}{dx}\left(x^2 \right) = 2x\quad \rightarrow \dfrac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}[/tex]
[tex]\dfrac{d}{dx}\left(\dfrac{p}{x}\right)=p \dfrac{d}{dx}\left(\dfrac{1}{x}\right) \quad\text{(since p is a constant)}[/tex]
[tex]\dfrac{d}{dx}\left(\dfrac{1}{x}\right) \\\\= \dfrac{d}{dx}\left(x^{-1}\right) \\\\= -1x^{-2}\\\\= -\dfrac{1}{x^2}[/tex]
Therefore
[tex]\dfrac{d}{dx}\left(\dfrac{p}{x}\right)= -p\left(\dfrac{1}{x^2}\right) = -\dfrac{p}{x^2}[/tex]
Putting all these individual terms together we get
[tex]f'(x) = 2x - \left(-\dfrac{p}{x}\right)[/tex]
[tex]f'(x) = 2x + \left\dfrac{p}{x^2}[/tex]
Setting [tex]f'(x) = 0[/tex]:
[tex]2x + \left\dfrac{p}{x^2} = 0[/tex]
[tex]\longrightarrow 2x = - \dfrac{p}{x^2}\\[/tex]
Multiplying both sides by [tex]x^2[/tex]:
[tex]2x \cdot x^2 = - p[/tex]
[tex]2x^3 = - p[/tex]
We are given the turning point T has an x-coordinate of 3
Plugging in [tex]x = 3[/tex] into [tex]2x^3 = - p[/tex] :
[tex]2(-3)^3 = - p[/tex]
[tex]2(-27) = -p[/tex]
[tex]-54 = - p[/tex]
[tex]p = 54[/tex]
Answer:
a) Value of [tex]p = 54[/tex]
Question 22(b)
We are given a curve
[tex]y = x^2 - \dfrac{16}{x}[/tex]
and we are told that the line [tex]y = k[/tex] is a tangent to this curve
The given line is a horizontal line intersecting the y-axis at y = k
The only point where it can be tangent to the line is at the turning point
If this line is a tangent to the curve then the line touches the curve at y = k and the slope of the curve at y = k is the same as the slope of the line
As in part (a) we can find the instantaneous slope at any point (x, y) on the curve by taking the first derivative.
[tex]\dfrac{dy}{dx} = \dfrac{d}{dx}\left(x^2 - \dfrac{16}{x}\right)\\\\[/tex]
[tex]> 2x - \left(-\dfrac{16}{x^2}\right) = 0\\\\= > 2x + \dfrac{16}{x^2} = 0\\\\= > 2x = -\dfrac{16}{x^2}\\\\= > 2x^3 = - 16\\\\x^3 = - 8\\[/tex]
[tex]x = \sqrt[3]{-8}\\\\x = -2[/tex]
Substituting for x = - 2 in
[tex]y = x^2 - \dfrac{16}{x}[/tex]
we get
[tex]y = (-2)^2 - \dfrac{16}{-2}\\\\y = 4 + 8\\\\y = 12\\[/tex]
Hence the value of k = 12
