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جستجو

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جستجوهای پرتکرار

میتونی لایو بذاری!

a) By sketching a suitable pair of graphs, show that the equation

${e^{2x}} = 14 - {x^2}$

has exactly two real roots.

b) Show by calculation that the positive root lies between 1.2 and 1.3.

c) Show that this root also satisfies the equation

$x = \frac{1}{2}ln\left( {14 - {x^2}} \right)$.

d) Use an iteration process based on the equation in part (c), with a suitable starting value, to find the root correct to 2 decimal places. Give the result of each step of the process to 4 decimal places.

پاسخ تشریحی :
نمایش پاسخ

a) Draw correct sketch of $y = {e^{2x}}$

Draw correct sketch of $y = 14 - {x^2}$

Indicate two real roots only from correct sketches

b) Consider sign of ${e^{2x}} + {x^2} - 14$ for 1.2 and 1.3 or equivalent

Justify conclusion with correct calculations $\left( {f\left( {1.2} \right) =  - 1.54,{\text{ }}f\left( {1.3} \right) = 1.15} \right)$

c) Confirm given answer $x = \frac{1}{2}ln\left( {14 - {x^2}} \right)$

d) Use the iteration process correctly at least once

Obtain final answer 1.26

Show sufficient iterations to 4 decimal places to justify answer or show a sign change in the interval $\left( {1.255,{\text{ }}1.256} \right)$

$[1.2 \to 1.2653 \to 1.2588 \to 1.2595{\text{ }};$

$1.25 \to 1.2604 \to 1.2593 \to 1.2594;$

$1.3 \to 1.2522 \to 1.2598 \to 1.2594]$

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