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A curve C has equation $y = e^x + x^4 + 8x + 5$ - Edexcel - A-Level Maths Pure - Question 3 - 2014 - Paper 6
Question 3
A curve C has equation $y = e^x + x^4 + 8x + 5$. (a) Show that the x coordinate of any turning point of C satisfies the equation $x^2 = 2 - e^{-x}$. (b) On the axe... show full transcript
Worked Solution & Example Answer:A curve C has equation $y = e^x + x^4 + 8x + 5$ - Edexcel - A-Level Maths Pure - Question 3 - 2014 - Paper 6
Step 1
Show that the x coordinate of any turning point of C satisfies the equation
Answer
To find the x-coordinate of any turning point, we need to determine where the derivative of the function equals zero:
- Differentiate the equation of curve C:
[ \frac{dy}{dx} = e^x + 4x^3 + 8 ] - Set ( \frac{dy}{dx} = 0 ):
[ e^x + 4x^3 + 8 = 0 ] - Rearranging, we get:
[ e^x = -4x^3 - 8 ]. - Taking the natural logarithm on both sides leads to the equation:
[ x^2 = 2 - e^{-x} ]
This shows the relationship the question asks for.
Step 2
On the axes given on page 5, sketch, on a single diagram, the curves with equations (i)
Answer
For the first curve (y = x^3):
- This curve passes through the origin (0, 0) and is symmetric about the origin.
- It will increase for all values of x, showing a typical S-shape.
For the second curve (y = 2 - e^{-x}):
- This curve begins at y = 2 when x=0 (crossing the y-axis) and approaches y = 2 as x approaches positive infinity.
- As x approaches negative infinity, the curve goes towards negative infinity.
- The graph has a horizontal asymptote at y = 2.
Make sure to label both curves and their key features.
Step 3
On your diagram give the coordinates of the points where each curve crosses the y-axis and state the equation of any asymptotes.
Answer
For the curve (y = x^3):
- It crosses the y-axis at the point (0, 0).
For the curve (y = 2 - e^{-x}):
- It crosses the y-axis at the point (0, 2).
- The horizontal asymptote is at y = 2.
Step 4
Explain how your diagram illustrates that the equation x^2 = 2 - e^{-x} has only one root.
Answer
From the sketch, observe that the curve (y = x^2) intersects with (y = 2 - e^{-x}) at one point. This shows that the equation has only one solution, as the two functions cross at exactly one point.
Step 5
Calculate the values of x_1 and x_2, giving your answers to 5 decimal places.
Answer
Using the iteration formula (x_{n+1} = (-2 - e^{-x_n})^{1/3}):
- Starting value: (x_0 = -1).
Calculate (x_1):
[x_1 = (-2 - e^{-(-1)})^{1/3} \approx -1.26376 ] - Now using (x_1) to find (x_2):
[x_2 = (-2 - e^{-(-1.26376)})^{1/3} \approx -1.26126 ]
Both values rounded to 5 decimal places are (x_1 = -1.26376) and (x_2 = -1.26126).
Step 6
Hence deduce the coordinates, to 2 decimal places, of the turning point of the curve C.
Answer
We have established that the turning point of the curve occurs around (x = -1.26). Substituting this x-value back into the original curve equation to find y:
[y = e^{-1.26} + (-1.26)^4 + 8(-1.26) + 5 \approx 4.48 ]
Thus, the turning point of curve C is approximately at the coordinates (-1.26, 4.48).