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Math  /  Calculus

Question(c) y=2x3ln⁡xy=2^{x^{3} \ln x}
The curve y=2x3+bx2+cxy=2 x^{3}+b x^{2}+c x has a turning point at (1,5)(1,5). (a) Find the values of bb and cc. (b) By using the second derivative test, determine whether the stationary point at (1,5)(1,5) is a minimum or maximum point.

Studdy Solution

STEP 1

1. A turning point occurs where the first derivative of a function is zero.
2. The second derivative test can determine the nature of a turning point.
3. The given curve is a polynomial, so derivatives will be straightforward to calculate.

STEP 2

1. Differentiate the curve y=2x3+bx2+cx y = 2x^3 + bx^2 + cx to find the first derivative.
2. Use the condition of the turning point at (1,5) (1,5) to set up equations for b b and c c .
3. Solve the equations to find the values of b b and c c .
4. Differentiate the curve again to find the second derivative.
5. Use the second derivative test to determine the nature of the turning point at (1,5) (1,5) .

STEP 3

Differentiate the curve y=2x3+bx2+cx y = 2x^3 + bx^2 + cx to find the first derivative:
y′=ddx(2x3+bx2+cx)=6x2+2bx+c y' = \frac{d}{dx}(2x^3 + bx^2 + cx) = 6x^2 + 2bx + c

STEP 4

Since there is a turning point at (1,5) (1,5) , the first derivative at x=1 x = 1 must be zero:
y′(1)=6(1)2+2b(1)+c=0 y'(1) = 6(1)^2 + 2b(1) + c = 0 6+2b+c=0 6 + 2b + c = 0
Additionally, since the point (1,5) (1,5) lies on the curve, substitute x=1 x = 1 and y=5 y = 5 into the original equation:
2(1)3+b(1)2+c(1)=5 2(1)^3 + b(1)^2 + c(1) = 5 2+b+c=5 2 + b + c = 5

STEP 5

Now solve the system of equations:
1. 6+2b+c=0 6 + 2b + c = 0
2. 2+b+c=5 2 + b + c = 5
Subtract the second equation from the first:
(6+2b+c)−(2+b+c)=0−5 (6 + 2b + c) - (2 + b + c) = 0 - 5 4+b=−5 4 + b = -5 b=−9 b = -9
Substitute b=−9 b = -9 into the second equation:
2−9+c=5 2 - 9 + c = 5 −7+c=5 -7 + c = 5 c=12 c = 12
Thus, b=−9 b = -9 and c=12 c = 12 .

STEP 6

Differentiate the first derivative to find the second derivative:
y′′=ddx(6x2+2bx+c)=12x+2b y'' = \frac{d}{dx}(6x^2 + 2bx + c) = 12x + 2b

STEP 7

Use the second derivative test to determine the nature of the turning point at (1,5) (1,5) :
Substitute x=1 x = 1 into the second derivative:
y′′(1)=12(1)+2(−9) y''(1) = 12(1) + 2(-9) y′′(1)=12−18 y''(1) = 12 - 18 y′′(1)=−6 y''(1) = -6
Since y′′(1)<0 y''(1) < 0 , the turning point at (1,5) (1,5) is a maximum point.
The values of b b and c c are b=−9 b = -9 and c=12 c = 12 , and the turning point at (1,5) (1,5) is a maximum point.

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