Applications of Derivatives Calculator

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Example

Created on 2024-06-20Asked by Ella Thompson (Solvelet student)

Find the critical points of the function

f(x) = x^3 - 3x^2 + 4

and determine their nature (local maxima, minima, or saddle points).

Solution

Step 1: Find the first derivative of the function

f(x) = x^3 - 3x^2 + 4

f'(x) = 3x^2 - 6x.

Step 2: Find the critical points by setting

f'(x) = 0

3x^2 - 6x = 0 \implies 3x(x - 2) = 0 \implies x = 0 \text{ or } x = 2.

Step 3: Determine the nature of the critical points using the second derivative. Find

f''(x)

f''(x) = 6x - 6.

Step 4: Evaluate

f''(x)

at the critical points:

f''(0) = 6(0) - 6 = -6 \quad (\text{negative, so } x = 0 \text{ is a local maximum}),

f''(2) = 6(2) - 6 = 6 \quad (\text{positive, so } x = 2 \text{ is a local minimum}).

Step 5: Conclusion. The critical points of

f(x) = x^3 - 3x^2 + 4

are

x = 0

(local maximum) and

x = 2

(local minimum). Solved on Solvelet with Basic AI Model

Some of the related questions asked by Amelia Young on Solvelet

1. Find the critical points of the function

f(x) = x^3 - 6x^2 + 9x + 5

2. Determine the function intervals is either increasing or decreasing

f(x) = 2x^3 - 9x^2 + 12x - 3

DefinitionIn more generally, derivatives can be applied to modelling and calculating variations in a distinct measure such as demand and supply, cost functions (economic derivatives), or rates of gradients of curves (which are essential for solving physics problems on motion), and optimization problems in engineering. They help in finding the points of maxima, minima and points of inflection. For an example: maximize the profit, set the profit function equal to its derivative then find the critical point, to use the second derivative test.

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