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).
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
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 ...
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.
find out the stationary point, saddle point, min and max
1) Let f(x,y)=25e^(-1/5x^2)-y^5+5y+3
a) Find all stationary points of the function f(x,y) and enter their coordinates by "" with at least 3 decimal places.
b)Let (xs,ys) be the saddle point of the function f(x,y). Calculate the following expression:
f(xs,ys)-(xs+ys) and enter the value with at least 3 dp.
c) what is(are) type of stationarity the other point(s)?
a) stationary points
b) let (xs,ys) be the saddle point of the function f(x,y), calculate the following expression f(xs,ys)-(xs+ys)
c)what type of stationary points (stationary, saddle, min, max).