`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.

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Textbook Question

Chapter 2, 2.4 - Problem 75 - Calculus of a Single Variable (10th Edition, Ron Larson).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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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` .

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