Find a function f such that { f '(x) = 3x^3 } and the line 3x + y = 0 is tangent to the graph...

Question:

Find a function f such that {eq}f '(x) = 3x^3 {/eq} and the line 3x + y = 0 is tangent to the graph of f.

Indefinite integration

This example illustrates how to use integration to find the Cartesian equation of a curve, when the curve must satisfy certain given properties which are pre-specified.

Answer and Explanation:

Firstly, since {eq}f '(x) = 3x^3 {/eq}, it follows that

{eq}\displaystyle f(x) = \int 3x^3\: dx = \frac{3}{4}x^4 + c. {/eq}

The given tangent line 3x + y = 0 has gradient -3.

But the gradient of the curve is given by f'(x), so at the point of contact of the tangent and the curve, we must have

{eq}f'(x) = 3x^3 = -3 \\ \Rightarrow x^3 = -1 \\ \Rightarrow x = \sqrt[3]{-1} = -1. {/eq}

But when x = -1, {eq}\displaystyle y = f(-1) = \frac{3}{4}(-1)^4 + c = \frac{3}{4} + c. {/eq}

So the equation of the tangent to the curve at x = -1 will have gradient -3 and y coordinate {eq}\displaystyle \frac{3}{4} + c. {/eq}

Then using {eq}y- y_1 = m(x - x_1) {/eq} the equation of the tangent at x = -1 is

{eq}\displaystyle y - (\frac{3}{4} + c) = -3(x - (-1)) \\ \displaystyle y + 3x = c - \frac{9}{4}. {/eq}

But we know that the tangent line here has equation y + 3x = 0.

It follows that

{eq}\displaystyle c - \frac{9}{4} = 0 \\ \Rightarrow \displaystyle c = \frac{9}{4}. {/eq}

Hence the required function is

{eq}\displaystyle f(x) = \frac{3}{4}x^4 + \frac{9}{4}. {/eq}


Learn more about this topic:

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Integration Problems in Calculus: Solutions & Examples

from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13
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