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I don't understand how to use Newton's method for this question.

Consider the function $f(x) = 3 − x − e^x$

(i) Find the equations of the tangent and the normal to the curve $y = f(x)$ at the point $P(0, 2)$.

(ii) Use Newton’s method in 2 steps to solve the equation $ 3 − x − e^x = 0$ for an approximation of a root near $x = 0$. Give your approximation value in 4 decimal places.

My working out

i) $y'= -1-e^x$

Sub in x = 0, to find the gradient for the equation of the tangent.

$y'= -2$

The tangent equation of the line is

$ y=mx + c$

$2 = -2(0) + c$

$c = 2 $

Therefore the tangent equation is

$y = -2x + 2$

Finding the equation of the tangent.

Since $ y' = -2$ for the line the normal equation's gradient is therefore

$y'= 1/2$

The equation of the normal is

$ y = mx + c$

$ 2 = 0.5(0) + c$

$ c = 2$

Therefore $ y = 0.5x + 2$

Maggie
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1 Answers1

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Under 1) you should not sub in $x=0$ but leave it with $y'=-1-e^x$. You start with $x_0=0, y_0=2, y'_0=-2$ Then $x_1=x_0-\frac {y_0}{y'_0}$. Intuitively you want to reduce $y_0$ to zero, so you slide down the tangent to where you should find a root. This gives you $x_1$ from which you find $y_1, y'_1$. Repeat.

Ross Millikan
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