An equation of the tangent line to the curve y=f(x)=x(4cosx−6sinx) at the point (3f(3)) is

y =?

Note: Please follow these instructions carefully:
(a) Express f(3) and f(3) in terms of . Simplify those expressions by evaluating the exact values of sin(3) and cos(3).
(b) Your answer should be in the form ax+m. When calculating a and m, approximate by 3.14.

y = x(4cosx-6sinx)

y(3) = 3(4cos3-6sin3) = -14.42

y' = -2(3x-2)cosx - 2(2x+3)sinx
y'(3) = 11.32

Now you have a point and a slope.
The equation of the line is cinchy.

THANK YOU SO MUCH!

To find the equation of the tangent line to the curve at the point (3f(3)), we will follow these steps:

(a) Express f(3) and f'(3) in terms of sin(3) and cos(3).

First, let's find f(3). We are given that f(x) = x(4cos(x) - 6sin(x)).
Evaluate f(3) by substituting x = 3 into the function:
f(3) = 3(4cos(3) - 6sin(3)).

To simplify the expression, we need to evaluate the exact values of sin(3) and cos(3).
Unfortunately, I am only able to provide an approximation based on the value of π (pi), which is approximately 3.14.

Using this approximation, you can calculate sin(3) and cos(3) as follows:
sin(3) ≈ sin(3.14) ≈ 0.00159,
cos(3) ≈ cos(3.14) ≈ -0.99999.

Now, let's substitute these approximated values back into f(3):
f(3) ≈ 3(4*(-0.99999) - 6*(0.00159)).

Simplifying the expression:
f(3) ≈ 3(-3.99996 - 0.00954).
f(3) ≈ 3(-4.0095).
f(3) ≈ -12.0285.

So, f(3) ≈ -12.0285.

(b) Now that we have f(3) ≈ -12.0285, we can use this value to find the slope of the tangent line, which is the derivative of f(x) evaluated at x = 3.

To find f'(x), we will differentiate f(x) = x(4cos(x) - 6sin(x)):
f'(x) = (4cos(x) - 6sin(x)) + x(-4sin(x) - 6cos(x)).

Evaluate f'(3) by substituting x = 3:
f'(3) = (4cos(3) - 6sin(3)) + 3(-4sin(3) - 6cos(3)).

Using the approximated values of sin(3) and cos(3) from before:
f'(3) ≈ (4*(-0.99999) - 6*(0.00159)) + 3(-4*(0.00159) - 6*(-0.99999)).

Simplifying the expression:
f'(3) ≈ (-3.99996 - 0.00954) + 3(-0.00636 + 5.99994).
f'(3) ≈ (-4.00950) + 3(5.99358).
f'(3) ≈ -4.00950 + 17.98074.
f'(3) ≈ 13.97124.

So, f'(3) ≈ 13.97124.

Now, we have the point (3f(3)), which is (3, -12.0285), and the slope of the tangent line, which is f'(3) ≈ 13.97124.

Using the point-slope form of a line, y - y1 = m(x - x1), we can substitute the values:
y - (-12.0285) = 13.97124(x - 3).

Simplifying the equation:
y + 12.0285 = 13.97124(x - 3).

Finally, we can rewrite the equation in the desired form ax + m:
y = 13.97124x - 41.91372 - 12.0285.

Simplifying further:
y = 13.97124x - 53.94222.

Therefore, the equation of the tangent line to the curve y = f(x) = x(4cos(x) - 6sin(x)) at the point (3f(3)) is y = 13.97124x - 53.94222 (approximated to two decimal places).