-8x 20 is a tangent to f(x) x^3 x^2 bx 18 at x=1 calculate the values of a and b
I tried the question but I failed
x^3+x^2+bx+18
-8x+20
f = x^3+x^2+bx+18
f' = 3x^2 + 2x + b
f'(1) = 3+2+b = 5+b
Since y = -8x+20 is tangent at x=1, it has the same slope as the curve there. So,
5+b = -8
b = -13
I have no idea where a comes into play, but maybe you can work it in and include it in the solution.
To solve this problem, we will first find the derivative of the function f(x), and then substitute x=1 into the equation of the tangent line. This will allow us to form two equations and solve for the unknowns a and b.
Let's start by finding the derivative of f(x). Since f(x) is given as x^3 + x^2 + bx + 18, we differentiate each term with respect to x:
f'(x) = 3x^2 + 2x + b
Since the given line -8x + 20 is a tangent to f(x) at x=1, we know that the slope of the tangent line is equal to the derivative of f(x) evaluated at x=1. So, we set up our first equation:
f'(1) = -8
By substituting x=1 into the derivative, we solve for the value of b:
3(1)^2 + 2(1) + b = -8
3 + 2 + b = -8
b = -13
Now that we have found the value of b, we can proceed to find the value of a. We substitute x=1 into the equation of f(x) and the equation of the tangent line, and set them equal to each other:
f(1) = -8(1) + 20
By substitute x=1 into f(x), we obtain:
1^3 + 1^2 + (-13)(1) + 18 = -8(1) + 20
1 + 1 - 13 + 18 = -8 + 20
7 = 12
Since 7 does not equal 12, this implies that there is no value that satisfies the equation. Therefore, there is no value for a that satisfies the given conditions.
To summarize:
a = No solution
b = -13