Respuesta :
Answer:
1) a = 2, b = 6
2) (i) a = 2, b = 3
(ii) (0, 0) and (-2, 8)
Step-by-step explanation:
Question 1
The gradient of the tangent line to f(x) = ax³ + bx² at point (x, y) is the derivative of the function at that point.
Differentiate function f(x):
[tex]f'(x)=3ax^2 + 2bx[/tex]
To find the expression for the gradient (m) of the tangent line at point (-1, 4), substitute x = -1 into f'(x):
[tex]m=f'(-1)\\\\m=3a(-1)^2+2b(-1)\\\\m=3a-2b[/tex]
Now, substitute the coordinates of point (-1, 4) into function f(x) and rearrange to isolate b:
[tex]a(-1)^3+b(-1)^2=4\\\\-a+b=4\\\\b=a+4[/tex]
Substitute the expression for b into the gradient equation so that the gradient (m) is in terms of a only:
[tex]m=3a-2(a+4)\\\\m=3a-2a-8\\\\m=a-8[/tex]
The gradient of the tangent line y = -6x - 2 is m = -6.
Substitute this into the equation m = a - 8 and solve for a:
[tex]a-8=-6\\\\a=2[/tex]
Now, substitute the found value of a into b = a + 4, and solve for b:
[tex]b = 2 + 4\\\\b=6[/tex]
Therefore, the values of a and b for which the line y = -6x - 2 is a tangent to the graph of f(x) = ax³ + bx² at point (-1, 4) are:
[tex]\Large\boxed{\boxed{a=2}}[/tex]
[tex]\Large\boxed{\boxed{b=6}}[/tex]
[tex]\dotfill[/tex]
Question 2
The gradient of the tangent line to y = ax³ + bx² at point (x, y) is equal to the derivative of the function at that point.
Differentiate the equation:
[tex]\dfrac{dy}{dx}=3ax^2+2bx[/tex]
Given that the gradient at point (1, 5) is 12, substitute x = 1 and dy/dx = 12 into the derivative equation:
[tex]3a(1)^2+2b(1)=12\\\\3a+2b=12[/tex]
Substitute the coordinates of point (1, 5) into the equation of the curve and rearrange to isolate b:
[tex]a(1)^3+b(1)^2=5\\\\a+b=5\\\\b=5-a[/tex]
Substitute the expression for b into the first equation and solve for a:
[tex]3a+2(5-a)=12\\\\3a+10-2a=12\\\\a+10=12\\\\a=2[/tex]
Substitute the value of a into the equation for b, and solve for b:
[tex]b = 5 - 2\\\\b=3[/tex]
Therefore, the values of a and b for which the gradient of the tangent line to the graph of f(x) = ax³ + bx² at point (1, 5) is 12 are:
[tex]\Large\boxed{\boxed{a=2}}[/tex]
[tex]\Large\boxed{\boxed{b=3}}[/tex]
So, the equation of the line is:
[tex]y=2x^3 + 3x^2[/tex]
The points on the curve at which the tangent is parallel to the x-axis are when the gradient of the tangent is zero.
Differentiate y = 2x³ + 3x²:
[tex]\dfrac{dy}{dx}=6x^2+6x[/tex]
Set dy/dx equal to zero and solve for x:
[tex]6x^2+6x=0\\\\6x(x+1)=0\\\\\\6x=0 \implies x=0\\\\x+1=0\implies x=-1[/tex]
Therefore, the x-coordinates of the points are x = 0 and x = -1.
To find the corresponding y-coordinates, substitute the x-coordinates into the equation of the curve, y = 2x³ + 3x²:
[tex]y=2(0)^3+3(0)^2\\\\y=0[/tex]
[tex]y=2(-1)^3+3(-1)^2\\\\y=1[/tex]
Therefore, the coordinates of the points on the curve where the tangent to the curve is parallel to the x-axis are:
[tex]\Large\boxed{\boxed{(0, 0)}}[/tex]
[tex]\Large\boxed{\boxed{(-1,1)}}[/tex]
