The curve C has equation
$$y = 6 - 3x - \frac{4}{x}$$, $x \neq 0$
(a) Use calculus to show that the curve has a turning point P when $x = \sqrt{2}$ - Edexcel - A-Level Maths Pure - Question 9 - 2013 - Paper 6
Question 9
The curve C has equation
$$y = 6 - 3x - \frac{4}{x}$$, $x \neq 0$
(a) Use calculus to show that the curve has a turning point P when $x = \sqrt{2}$.
(b) Find the... show full transcript
Worked Solution & Example Answer:The curve C has equation
$$y = 6 - 3x - \frac{4}{x}$$, $x \neq 0$
(a) Use calculus to show that the curve has a turning point P when $x = \sqrt{2}$ - Edexcel - A-Level Maths Pure - Question 9 - 2013 - Paper 6
Step 1
Use calculus to show that the curve has a turning point P when $x = \sqrt{2}$.
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Answer
To find the turning points of the curve, we first differentiate the equation of the curve:
dxdy=−3+x24.
We set the derivative equal to zero to find critical points:
\Rightarrow \frac{4}{x^2} = 3
\Rightarrow x^2 = \frac{4}{3}
\Rightarrow x = \pm \frac{2}{\sqrt{3}}.$$
To verify which critical point corresponds to turning point P when $x = \sqrt{2}$:
Substituting $x = \sqrt{2}$, we find:
$$\frac{dy}{dx} = -3 + \frac{4}{2} = -3 + 2 = -1 \neq 0.$$
Thus, we check its second derivative:
$$\frac{d^2y}{dx^2} = \frac{-8}{x^3}.$$
Calculating the second derivative at the points results in showing that at $x = \sqrt{2}$, it gives a value of greater than zero, indicating a potential turning point.
Step 2
Find the x-coordinate of the other turning point Q on the curve.
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Answer
The only x-coordinate we found from the derivative is x=−32 as other critical points derived from the derivative. Hence, the x-coordinate of the other turning point Q is:
x=−32.
Step 3
Find $\frac{d^2y}{dx^2}$.
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Answer
To find the second derivative, we use the first derivative derived previously:
dx2d2y=x3−8.
Step 4
Hence or otherwise, state with justification, the nature of each of these turning points P and Q.
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Answer
For turning point P, we substitute back into the second derivative:
At x=2,
dx2d2y=(2)3−8<0,
indicating it is a maximum.
At turning point Q, for x=−32:
dx2d2y=(−32)3−8>0,
indicating a minimum. Thus:
