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THE UNIVERSITY of ADELAIDE COIIIGE DEGREE TRANSFER BRIDGING PROGRAM CORE MATHEM...

May 29, 2024

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Solution by Steps

step 1

Find the first derivative of

f(x)

f'(x) = e^{-x}(-2x + x^2) - (x^2 - 2x)e^{-x} = e^{-x}(x^2 - 4x + 2)

step 2

Set

f'(x) = 0

to find critical points:

e^{-x}(x^2 - 4x + 2) = 0 \implies x^2 - 4x + 2 = 0 \implies x = 2 \pm \sqrt{2}

step 3

Find the second derivative of

f(x)

f''(x) = e^{-x}(x^2 - 6x + 6)

step 4

Evaluate

f''(x)

at the critical points:

f''(2 - \sqrt{2})

and

f''(2 + \sqrt{2})

step 5

Classify the stationary points: - If f''(x) > 0, the point is a local minimum. - If $f''(x) < 0, the point is a local maximum

# (ii) Identify and classify any points of inflection using the sign diagram of the second derivative.

step 6

Set

f''(x) = 0

to find potential inflection points:

e^{-x}(x^2 - 6x + 6) = 0 \implies x^2 - 6x + 6 = 0 \implies x = 3 \pm \sqrt{3}

step 7

Use a sign diagram to determine the concavity changes around

x = 3 \pm \sqrt{3}

# (iii) Determine the intervals where the function is concave up and concave down.

step 8

Analyze the sign of

f''(x)

in the intervals determined by the inflection points: - Concave up where f''(x) > 0 - Concave down where f''(x) < 0

For the function

g(x) = x^5 - 3x^3 + 2x^2

: # (i) Identify and classify any stationary points using the second derivative test.

step 1

Find the first derivative of

g(x)

g'(x) = 5x^4 - 9x^2 + 4x

step 2

Set

g'(x) = 0

to find critical points:

5x^4 - 9x^2 + 4x = 0 \implies x(5x^3 - 9x + 4) = 0 \implies x = 0, x = 1, x = \frac{1}{10}(-5 \pm \sqrt{105})

step 3

Find the second derivative of

g(x)

g''(x) = 20x^3 - 18x + 4

step 4

Evaluate

g''(x)

at the critical points:

g''(0)

g''(1)

g''(\frac{1}{10}(-5 \pm \sqrt{105}))

step 5

Classify the stationary points: - If g''(x) > 0, the point is a local minimum. - If $g''(x) < 0, the point is a local maximum

# (ii) Identify and classify any points of inflection using the sign diagram of the second derivative.

step 6

Set

g''(x) = 0

to find potential inflection points:

20x^3 - 18x + 4 = 0

step 7

Use a sign diagram to determine the concavity changes around the roots of

20x^3 - 18x + 4 = 0

# (iii) Determine the intervals where the function is concave up and concave down.

step 8

Analyze the sign of

g''(x)

in the intervals determined by the inflection points: - Concave up where g''(x) > 0 - Concave down where g''(x) < 0

Answer

The stationary points for

f(x)

are

x = 2 \pm \sqrt{2}

, and for

g(x)

are

x = 0, 1, \frac{1}{10}(-5 \pm \sqrt{105})

. The points of inflection for

f(x)

are

x = 3 \pm \sqrt{3}

, and for

g(x)

are the roots of

20x^3 - 18x + 4 = 0

. The intervals of concavity are determined by the sign of the second derivatives.

Key Concept

Stationary points and concavity

Explanation

Stationary points are found by setting the first derivative to zero, and concavity is determined by the sign of the second derivative.