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 axes given on page 5, sketch, on a single diagram, the curves with equations (i) $y = x^3$. (ii) $y = 2 - e^{-x}$. On your diagram give the coordinates of the points where each curve crosses the $y$-axis and state the equation of any asymptotes. (c) Explain how your diagram illustrates that the equation $x^2 = 2 - e^{-x}$ has only one root. The iteration formula $$x_{n+1} = (-2 - e^{-x_n})^{rac{1}{3}}, \, x_0 = -1$$ can be used to find an approximate value for this root. (d) Calculate the values of $x_1$ and $x_2$, giving your answers to 5 decimal places. (e) Hence deduce the coordinates, to 2 decimal places, of the turning point of the curve C.

A-Level Edexcel Maths Pure Trigonometric Equations: A curve C has equation $y = e^x

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 ... 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

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Answer

To find the turning points, we first differentiate the function:

$\frac{dy}{dx} = e^x + 4x^3 + 8$

Setting the derivative equal to zero gives:

$e^x + 4x^3 + 8 = 0$

Rearranging this leads to:

$e^x = -4x^3 - 8$

Therefore, squaring both sides results in:

$x^2 = 2 - e^{-x}$

Step 2

(i) $y = x^3$.

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Answer

The curve $y = x^3$ should start at the origin (0,0) and pass through Quadrants 1 and 3. It has a positive gradient there.

Step 3

(ii) $y = 2 - e^{-x}$.

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Answer

For the curve $y = 2 - e^{-x}$ , it crosses the $y$ -axis at (0, 1) since $e^{0} = 1$ . The asymptote is at $y = 2$ .

Step 4

Explain how your diagram illustrates that the equation $x^2 = 2 - e^{-x}$ has only one root.

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Answer

The sketch of $y = x^3$ and $y = 2 - e^{-x}$ shows that these two curves intersect at one point on the graph, which indicates there is only one turning point where the values of $y = x^3$ and $y = 2 - e^{-x}$ are equal.

Step 5

Calculate the values of $x_1$ and $x_2$, giving your answers to 5 decimal places.

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Answer

Using the iteration formula:

$x_1 = (-2 - e^{-(-1)})^{1/3}$ This results in

$x_1 \approx -1.26376$

Calculating $x_2$ :

$x_2 = (-2 - e^{-(-1.26376)})^{1/3}$ This results in

$x_2 \approx -1.26126$

Step 6

Hence deduce the coordinates, to 2 decimal places, of the turning point of the curve C.

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Answer

The turning point of the curve C is therefore approximately at the coordinates $(-1.26, y)$ where $y = e^{-1.26} + (-1.26)^4 + 8(-1.26) + 5$ . Calculating this, we find:

$y \approx 3.35$

Thus, the turning point is $(-1.26, 3.35)$ .

A curve C has equation $y = e^x + x^4 + 8x + 5$ - Edexcel - A-Level Maths Pure - Question 3 - 2014 - Paper 6

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

Show that the $x$ coordinate of any turning point of C satisfies the equation

(i) $y = x^3$.

(ii) $y = 2 - e^{-x}$.

Explain how your diagram illustrates that the equation $x^2 = 2 - e^{-x}$ has only one root.

Calculate the values of $x_1$ and $x_2$, giving your answers to 5 decimal places.

Hence deduce the coordinates, to 2 decimal places, of the turning point of the curve C.

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