Geo. App of Calculus - Types of stationary points (1 Viewer)

[pc4pc]

Hi there. the teacher wasn't available on this day.. and i think we're still expected to complete this exercise. help appreciated.

A curve has f'(x) = x(x+1). For what x values does the curve have turning points? What type are they?

also,

For a certain curve,
dy/dx = (x-1)^2(x-2). Find the x values of its turning points and determine their nature.

For both questions.. i keep getting 0 for either the LHS or RHS to find out the type.

 

[watatank]

A curve has f'(x) = x(x+1). For what x values does the curve have turning points? What type are they?

ok stationary points when f'(x) = 0 ie where x(x+1) = 0
x = 0, 1

to find the nature do the second derivative
f'(x) = x(x+1)
f'(x) = x^2 + x

f"(x) = 2x + 1

Test for x = 0,1

f"(0) = 2(0) + 1 = 1 which is >0 so its concave up, relative minimum

f"(1) = 2(1) + 1 = 3 which is >0 so its concave up, relative minimum

 

[watatank]

For a certain curve,
dy/dx = (x-1)^2(x-2). Find the x values of its turning points and determine their nature.

For both questions.. i keep getting 0 for either the LHS or RHS to find out the type.

Same as above, dy/dx = 0 to find the x value of the turning point
x = 1, 2

Find the derivative of that derivative, first expand it out...

(x-1)^2(x-2) = x^3 - 3x + 2 (cbf doin the working out)

d/dx (x^3 - 3x + 2) = 3x^2 - 3

sub 1, 2 for x

when x = 1, second derivative = 0
when x = 2 second derivative > 0, concave up, relative minimum

as you dont know the nature of the stat point straight away for x = 1 you have to do the gradient test, test the points in f '(x)

when x = 0.5 it is a positive gradient /
when x = 1 it is zero _
when x = 1.5 the gradient is positive /

So its a point of inflexion (vertical point)

Stat points: at x = 2, relative min
x = 1, vertical point of inflexion