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1 (i) Determine, algebraically, that

f(x)=\left(x^{2}+2 x+2\right) e^{-x}

and...

May 29, 2024

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Solution by Steps

step 1

Set

f(x)

equal to

g(x)

\left(x^{2}+2 x+2\right) e^{-x} = e^{-x}

step 2

Divide both sides by

e^{-x}

x^{2}+2 x+2 = 1

step 3

Simplify the equation:

x^{2}+2 x+1 = 0

step 4

Factorize the quadratic equation:

(x+1)^{2} = 0

step 5

Solve for

x

x = -1

step 6

Verify the solution by substituting

x = -1

back into

f(x)

and

g(x)

f(-1) = g(-1) = e^{-(-1)} = e

Answer

The functions

f(x)

and

g(x)

intersect at only one point,

x = -1

Key Concept

Intersection of functions

Explanation

To find the intersection points of two functions, set them equal to each other and solve for the variable.

Question 2: Determine that the slope of the tangent to the ellipse

3 x^{2}-2 x y+y^{2}=1

is given by the expression

\frac{d y}{d x}=\frac{3 x-y}{x-y}

step 1

Differentiate the ellipse equation implicitly with respect to

x

6x - 2y \frac{dy}{dx} - 2x \frac{dy}{dx} + 2y \frac{dy}{dx} = 0

step 2

Simplify the differentiated equation:

6x - 2y \frac{dy}{dx} = 0

step 3

Solve for

\frac{dy}{dx}

\frac{dy}{dx} = \frac{3x - y}{x - y}

Answer

The slope of the tangent to the ellipse is

\frac{d y}{d x}=\frac{3 x-y}{x-y}

Key Concept

Implicit differentiation

Explanation

Implicit differentiation is used to find the derivative of a function defined by an equation involving both

x

and

y

Question 3: Determine the equations of the tangents to the ellipse that have slope 1.

step 1

Set the slope

\frac{dy}{dx}

equal to 1:

\frac{3x - y}{x - y} = 1

step 2

Solve for

y

3x - y = x - y \implies 2x = 0 \implies x = 0

step 3

Substitute

x = 0

into the ellipse equation to find

y

3(0)^2 - 2(0)y + y^2 = 1 \implies y^2 = 1 \implies y = \pm 1

step 4

Write the equations of the tangents:

y = 1

and

y = -1

Answer

The equations of the tangents to the ellipse with slope 1 are

y = 1

and

y = -1

Key Concept

Tangent line to an ellipse

Explanation

To find the tangent lines with a specific slope, set the derivative equal to the slope and solve for the points of tangency.

Question 4: Determine the equations of the normals to the ellipse that have slope

\frac{1}{2}

step 1

Set the slope

\frac{dy}{dx}

equal to

\frac{1}{2}

\frac{3x - y}{x - y} = \frac{1}{2}

step 2

Solve for

y

2(3x - y) = x - y \implies 6x - 2y = x - y \implies 5x = y

step 3

Substitute

y = 5x

into the ellipse equation:

3x^2 - 2x(5x) + (5x)^2 = 1 \implies 3x^2 - 10x^2 + 25x^2 = 1 \implies 18x^2 = 1 \implies x^2 = \frac{1}{18} \implies x = \pm \frac{1}{\sqrt{18}}

step 4

Find corresponding

y

values:

y = 5x \implies y = \pm \frac{5}{\sqrt{18}}

step 5

Write the equations of the normals:

y - \frac{5}{\sqrt{18}} = -2(x - \frac{1}{\sqrt{18}})

and

y + \frac{5}{\sqrt{18}} = -2(x + \frac{1}{\sqrt{18}})

Answer

The equations of the normals to the ellipse with slope

\frac{1}{2}

are

y - \frac{5}{\sqrt{18}} = -2(x - \frac{1}{\sqrt{18}})

and

y + \frac{5}{\sqrt{18}} = -2(x + \frac{1}{\sqrt{18}})

Key Concept

Normal line to an ellipse

Explanation

To find the normal lines with a specific slope, set the derivative equal to the negative reciprocal of the slope and solve for the points of normality.