# Stationary point

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This question concerns the function f(x)=x^3 +3x^2 −24x+40.

(a) Find the stationary points of this function.

(b) (i) Using the strategy to apply the First Derivative Test, classify the left-hand stationary point found in part (a).

(ii) Using the Second Derivative Test, classify the right-hand stationary point found in part (a).

(c) Find the y-coordinate of each of the stationary points on the graph of the function f (x), and also evaluate f (0).

(d) Hence draw a rough sketch of the graph of the function f(x).

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##### Solution Summary

For given function, stationary point is obtained. Followed by using first and second derivatives, left hand and right hand are classified. Finally y co-ordinate of the stationary point is obtained and the functions is drawn.

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(a)

f(x) = x^3 +3x^2 −24x+40

df/dx = 3x^2 + 6x - 24

For stationary points,

df/dx = 0

=> 3x^2 + 6x - 24 = 0

=> x^2 + 2x - 8 = 0

=> x^2 + 4x - 2x - 8 = 0

=> x (x + 4) - 2 (x + 4) = 0

=> (x+4) (x - 2) = 0

=> x = -4, 2

Hence, stationary points are: -4, 2

(b)(i)

First Derivative test: Find the slopes of the function near the stationary points (both the sides).

Left side of x = -4, x = -4-h, where h is a positive small fractional (<1) number

df/dx|(x : -4 -h) = 3(-4-h)^2 + 6(-4-h) - 24

= 3(16 ...

###### Education

- BEng, Allahabad University, India
- MSc , Pune University, India
- PhD (IP), Pune University, India

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