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A curve C has equation y = x^2 e^x - Edexcel - A-Level Maths Pure - Question 4 - 2007 - Paper 5
Question 4
A curve C has equation y = x^2 e^x. (a) Find \( \frac{dy}{dx} \) using the product rule for differentiation. (b) Hence find the coordinates of the turning points ... show full transcript
Worked Solution & Example Answer:A curve C has equation y = x^2 e^x - Edexcel - A-Level Maths Pure - Question 4 - 2007 - Paper 5
Step 1
Find \( \frac{dy}{dx} \) using the product rule for differentiation.
Answer
To differentiate the function ( y = x^2 e^x ), we will use the product rule, which states that if ( u = x^2 ) and ( v = e^x ), then:
[ \frac{dy}{dx} = \frac{du}{dx} v + u \frac{dv}{dx} ]
Calculating ( \frac{du}{dx} ) and ( \frac{dv}{dx} ):
- ( \frac{du}{dx} = 2x )
- ( \frac{dv}{dx} = e^x )
Substituting these into the product rule gives:
[ \frac{dy}{dx} = (2x)(e^x) + (x^2)(e^x) = e^x(2x + x^2) ]
Step 2
Hence find the coordinates of the turning points of C.
Answer
To find the turning points, we set ( \frac{dy}{dx} = 0 ):
[ e^x(2x + x^2) = 0 ]
Since ( e^x \neq 0 ) for any x, we solve:
[ 2x + x^2 = 0 ]
Factoring out ( x ):
[ x(2 + x) = 0 ]
This gives us two solutions:
- ( x = 0 )
- ( x = -2 )
Now, we can find the corresponding y-coordinates:
- For ( x = 0 ): ( y = 0^2 e^0 = 0 ), so the point is ( (0, 0) ).
- For ( x = -2 ): ( y = (-2)^2 e^{-2} = 4 e^{-2} ), so the point is ( (-2, 4 e^{-2}) ).
Step 3
Find \( \frac{d^2y}{dx^2} \).
Answer
To find the second derivative, we will differentiate ( \frac{dy}{dx} = e^x(2x + x^2) ) again using the product rule:
[ \frac{d^2y}{dx^2} = \frac{d}{dx}[e^x] (2x + x^2) + e^x \frac{d}{dx}(2x + x^2) ]
Calculating each part:
- ( \frac{d}{dx}[e^x] = e^x )
- ( \frac{d}{dx}(2x + x^2) = 2 + 2x )
Substituting these values gives: [ \frac{d^2y}{dx^2} = e^x(2x + x^2) + e^x(2 + 2x) = e^x(4x + x^2 + 2) ]
Step 4
Determine the nature of each turning point of the curve C.
Answer
To determine the nature of the turning points, we evaluate ( \frac{d^2y}{dx^2} ) at the turning points found in part (b):
-
At ( x = 0 ): [ \frac{d^2y}{dx^2} = e^0(4(0) + 0^2 + 2) = 2 > 0 ] Thus, the point ( (0, 0) ) is a local minimum.
-
At ( x = -2 ): [ \frac{d^2y}{dx^2} = e^{-2}(4(-2) + (-2)^2 + 2) = e^{-2}(-8 + 4 + 2) = e^{-2}(-2) < 0 ] Thus, the point ( (-2, 4 e^{-2}) ) is a local maximum.