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Chapter 15: Problem 342

Where are the maxima and minima of \(\mathrm{y}=3 \mathrm{x}^{3}+4 \mathrm{x}+7 ?\)

Short Answer

Expert verified
There are no maxima or minima for the function \(y = 3x^3 + 4x + 7\), as there are no critical points found after analyzing its first and second derivatives.

Step by step solution

01

Find the first derivative

To find the first derivative of the function, apply the power rule of differentiation to each term in the expression. The power rule states that if \(y = ax^n\), then the derivative, \(y'\), is given by \(y' = n(ax^{n-1})\). For the given function \(y = 3x^3 + 4x + 7\), we have: \(y' = \frac{d}{dx}(3x^3) + \frac{d}{dx}(4x) + \frac{d}{dx}(7) = 9x^2 + 4\)
02

Find the critical points

To find the critical points, set the first derivative equal to zero and solve for \(x\). \(9x^2 + 4 = 0\) To solve this equation, rearrange and divide by 9: \(x^2 = -\frac{4}{9}\) From this equation, we find that there are no real solutions of x. There are no critical points.
03

Analyze the second derivative

To determine the nature of the critical points, we need to compute the second derivative of the function and evaluate it at the critical points. However, since we have concluded that there are no critical points, there are no maxima or minima. For completeness, find the second derivative as follows: \(y'' = \frac{d^2}{dx^2}(3x^3 + 4x + 7) = \frac{d}{dx}(9x^2 + 4) = 18x\) In Conclusion, the function \(y = 3x^3 + 4x + 7\) has no maxima or minima, as there are no critical points.

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