Photo AI

The curve with equation $y = f(x) = 3xe^x - 1$ has a turning point $P$ - Edexcel - A-Level Maths Pure - Question 7 - 2009 - Paper 2

Question icon

Question 7

The-curve-with-equation-$y-=-f(x)-=-3xe^x---1$-has-a-turning-point-$P$-Edexcel-A-Level Maths Pure-Question 7-2009-Paper 2.png

The curve with equation $y = f(x) = 3xe^x - 1$ has a turning point $P$. (a) Find the exact coordinates of $P$. (b) The equation $f(x) = 0$ has a root between $x = ... show full transcript

Worked Solution & Example Answer:The curve with equation $y = f(x) = 3xe^x - 1$ has a turning point $P$ - Edexcel - A-Level Maths Pure - Question 7 - 2009 - Paper 2

Step 1

Find the exact coordinates of P.

96%

114 rated

Answer

To find the coordinates of the turning point PP, we first determine where the derivative of f(x)f(x) is zero.

Calculating the first derivative:

f(x)=3ex+3xexf'(x) = 3e^x + 3xe^x

Setting this to zero:

3ex(1+x)=03e^x(1 + x) = 0

This yields:

eq 0$$ Thus, we can solve: $$1 + x = 0 \implies x = -1$$ We substitute $x = -1$ back into the original function to find $y$: $$f(-1) = 3(-1)e^{-1} - 1 = -\frac{3}{e} - 1$$ Thus, the coordinates of point $P$ are $(-1, -\frac{3}{e} - 1)$.

Step 2

Use the iterative formula with $x_0 = 0.25$ to find, to 4 decimal places, the values of $x_1$, $x_2$, and $x_3$.

99%

104 rated

Answer

Using the iterative formula:

xn+1=13exnx_{n+1} = \frac{1}{3} e^{x_n}

Step 1: Using x0=0.25x_0 = 0.25:

x1=13e0.250.2596x_1 = \frac{1}{3} e^{0.25} \approx 0.2596

Step 2: Using x1=0.2596x_1 = 0.2596:

x2=13e0.25960.2571x_2 = \frac{1}{3} e^{0.2596} \approx 0.2571

Step 3: Using x2=0.2571x_2 = 0.2571:

x3=13e0.25710.2578x_3 = \frac{1}{3} e^{0.2571} \approx 0.2578

Step 3

By choosing a suitable interval, show that a root of $f(x) = 0$ is $x = 0.2576$ correct to 4 decimal places.

96%

101 rated

Answer

To demonstrate that a root lies within [0.2575,0.25765][0.2575, 0.25765], we evaluate:

f(0.2575)0.000109f(0.2575) \approx 0.000109 f(0.25765)0.000379f(0.25765) \approx -0.000379

Since f(0.2575)>0f(0.2575) > 0 and f(0.25765)<0f(0.25765) < 0, by the Intermediate Value Theorem, there exists at least one root in the interval (0.2575,0.25765)(0.2575, 0.25765). Thus, we conclude that x=0.2576x = 0.2576 is correct to 4 decimal places.

Join the A-Level students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;