`y = (4x^3 + 3)^2, (-1,1)` (a) Find an equation of the tangent line to the graph of f at the given point, (b) use a graphing utility to graph the function and its tangent line at the point,...

`y = (4x^3 + 3)^2, (-1,1)` (a) Find an equation of the tangent line to the graph of f at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of the graphing utility to confirm your results.

Expert Answers
sciencesolve eNotes educator| Certified Educator

You need to find the equation of the tangent line to the given curve, at the point (-1,1), using the formula:

`f(x) - f(-1) = f'(-1)( - (-1))`

You need to put y = f(x) and you need to notice that f(-1) = 1.

You need to evaluate the derivative of the given function, using chain rule, such that:

`f'(x) = ((4x^3 + 3)^2)' => f'(x) = 2(4x^3 + 3)*(4x^3 + 3)'`

`f'(x) = 2(4x^3 + 3)*(12x^2)`

You need to evaluate f'(x) at x = -1, hence, you need to replace -1 for x in equation of derivative:

`f'(-1) = 2(4(-1)^3 + 3)*(12(-1)^2)`

`f'(-1) = 24(-4 + 3) => f'(-1) = -24`

You need to replace the values into equation of tangent line, such that:

`f(x) - 1 = -24(x + 1) => y = -24x - 24 + 1 => y = -24x - 23`

Hence, evaluating the equation of the tangent line to the given curve, at the given point, yields `y = -24x - 23` .