I have been struggling with this problem I cannot get the right answer. Please give me the steps to solve this problem.
Let f(x)= (2x-9)(x^2 + 6)
(A) Find the equation of the line tangent to the graph of f(x) at x=3
(B) Find the values(s) of x where the tangent line is horizontal.
I would expand the function first, then find the derivative.
f(x) = 2x^3 + 12x - 9x^2 - 54
f '(x) = 6x^2 + 12 - 18x
when x = 3
f(3) = -3(15) = -45
f '(3) = 6(9) + 12 - 18(3) = 12
A) so you have a slope of 12 and a point (3,-45)
Use y = mx + b to find the tangent equation
B) set 6x^2 - 18x + 12 = 0
and solve
Hint: divide both sides by 6, then it factors
Identify the variable, constant, and coefficient of the expression: –10k + 15
To solve these questions, we can use calculus techniques. We will need to compute the derivative of the function to find the slope of the tangent line at a given point.
(A) Find the equation of the line tangent to the graph of f(x) at x=3:
1. Calculate the derivative of f(x). The derivative of f(x) is found by applying the product rule and differentiating each term separately:
f'(x) = (2x - 9)(d/dx)(x^2 + 6) + (x^2 + 6)(d/dx)(2x - 9)
2. Simplify the derivative using the power rule and the constant rule:
f'(x) = (2x - 9)(2x) + (x^2 + 6)(2)
3. Evaluate f'(x) at x = 3 to find the slope of the tangent line:
f'(3) = (2(3) - 9)(2(3)) + (3^2 + 6)(2)
4. Simplify f'(3) to find the slope.
5. Once you have the slope at the point x = 3, you can use the point-slope formula to find the equation of the tangent line:
y - y1 = m(x - x1)
Plug in the values of the point (x1, y1) as (3, f(3)) and the slope m calculated in the previous step.
(B) Find the value(s) of x where the tangent line is horizontal:
1. The equation of a horizontal line is y = constant. To find where the tangent line is horizontal, we need to set the derivative equal to 0 and solve for x.
2. Set f'(x) = 0 and solve for x to find the points where the tangent line is horizontal.
Let me know if you need further assistance with any step.