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Chapter 7: Problem 54

Determine the point(s), if any, at which the graph of the function has a horizontal tangent line. $$ y=x^{3}+3 x^{2} $$

Short Answer

Expert verified
The function \(y = x^3 + 3x^2\) has horizontal tangents at the points where \(x = 0\) and \(x = -2\).

Step by step solution

01

Compute the derivative

The derivative of the function \(y = x^3 + 3x^2\) can be obtained using the power rule which says that the derivative of \(x^n\) is \(nx^{n-1}\). Applying this rule, the derivative \(y' = 3x^2 + 6x\).
02

Set the derivative equal to zero

Set the derivative equal to zero and solve for x to find the values of x where there is a horizontal tangent line to the function. This gives the equation \(3x^2 + 6x = 0\).
03

Solve for x

Factor out a 3x, leaving the equation in the form \(3x(x + 2) = 0\). Setting each factor equal to zero gives the solutions \(x = 0\) and \(x = -2\). These are the x-values where the function has a horizontal tangent.

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Most popular questions from this chapter

Find the derivative of the function. State which differentiation rule(s) you used to find the derivative. $$ y=\frac{1}{\sqrt{x+2}} $$

You deposit in an account with an annual interest rate of \(r\) (in decimal form) compounded monthly. At the end of 5 years, the balance is \(A=1000\left(1+\frac{r}{12}\right)^{60}\) Find the rates of change of \(A\) with respect to \(r\) when (a) \(r=0.08\), (b) \(r=0.10\), and (c) \(r=0.12\).

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