Finding stationary points (1 Viewer)

[reggie7]

Could someone explain how you figure out these questions. I'm stuck

From easiest to hardest:

1. Find the domain over which the curve y = x^3 + 12x^2 + 45x - 30 is increasing.

2. Find the stationary points on the curve y = (3x - 1) (x - 2)^4

3. Differentiate y = x√x + 1. Hence find the stationary point on the curve, giving the exact value.

 

[gurmies]

1

y = x^3 + 12x^2 + 45x - 30

For increasing values, dy/dx > 0

dy/dx = 3x^2 + 24x + 45

3x^2 + 24x + 45 > 0

x^2 + 8x + 15 > 0

Hence, x < -5 or/ x > -3

2

y = (3x-1)(x-2)^4

dy/dx = (x-2)^4 (3) + (3x-1) x 4(x-2)^3

= (x-2)^3 [3(x-2) + 4(3x-1)

= (x-2)^3[3x - 6 + 12x - 4]

= (x-2)^3[15x-10]

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

For S.Ps, dy/dx = 0

5(3x-2)(x-2)^3 = 0

x = 2/3 or/ x = 2 [2/3, 256/81]

y = 256/81 or/ y = 0 [2, 0]

Since you didn't specify whether you wanted turning point or point of inflexion, I will not check the second derivative.

3

y = x√x + 1

y = x^(3/2) + 1

dy/dx = 3/2 (x)^(1/2)

= (3√x)/2

At S.Ps, dy/dx = 0

(3√x)/2 = 0

3√x = 0

x = 0

y = 1

(0,1) is the stationary point

 

[reggie7]

Thanks, Although I still dont get question 3, you lose me when you change it to the 2nd line lol.

 

[Timothy.Siu]

Thanks, Although I still dont get question 3, you lose me when you change it to the 2nd line lol.

xrootx=x^3/2
xrootx=root(x^3)=x^3/2
or x(x^1/2)=x^3/2

 

[gurmies]

Think of it in terms of the index laws:

(x^n)(x^m) = x^(n+m)

Similarly, (x)(x)^(1/2) = x^(1 + 1/2) = x^(3/2)

 

[reggie7]

sorry, question 3 was x√(x + 1) ...to make it more clear.

Could someone please re do question 3 please?

 

[gurmies]

OH, pardon me then =)

y = x√(x + 1)

y = x(x+1)^(1/2)

dy/dx = √(x + 1) + x . 1/2 . 1/√(x + 1)

= [2(x+1) + x] /2√(x + 1)

= (2x + 2 + x)/2√(x + 1)

= (3x + 2)/2√(x + 1)

At S.P's, dy/dx = 0

Hence, 3x + 2 = 0

x = -2/3

y = (-2√3)/9

 

[foram]

 

[reggie7]

thanks gurmies and foram.

at the end of question 1:

x^2 + 8x + 15 > 0
Hence, x < -5 or/ x > -3

I always get confused with changing the sign, so when you have (x +5) (x + 3) > 0, who bring the -5 over which changes the sign right? and then you bring the -3 over which changes it back to >....?

And with question 2, how come there are two x values? This is my working out:

dy/dx = (3x - 1) 4(x - 2)^3 + 3(x - 2)^4
3x - 1 = -3(x - 2)^4 ÷ 4(x - 2)^3
3x - 1 = -3(x - 2) ÷ 4
3x - 1 = -3x + 6 ÷ 4
4(3x - 1) = -3x + 6
15x = 10
x = 2/3
y= 3r13r81

 

[bored of sc]

2. Find the stationary points on the curve y = (3x - 1) (x - 2)^4

dy/dx = (3x-1)4(x-2)³ + 3(x-2)⁴
= (x-2)³(12x-4+3x-6)
= (x-2)³(15x-10)

Let dy/dx = 0 for stationary points.

(x-2)³(15x-10) = 0
x = 2, 2/3

Remember to treat each bracket as a separate equation equal to zero when finding the x-values of the stationary points. That's why there's two x-values.