Problem 54
Question
Determine the point(s), if any, at which the graph of the function has a horizontal tangent line. $$ y=x^{3}+3 x^{2} $$
Step-by-Step Solution
Verified Answer
The function \(y = x^3 + 3x^2\) has horizontal tangents at the points where \(x = 0\) and \(x = -2\).
1Step 1: 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 \(y' = 3x^2 + 6x\).
2Step 2: 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 \(3x^2 + 6x = 0\).
3Step 3: 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.
Other exercises in this chapter
Problem 54
Find the derivative of the function. State which differentiation rule(s) you used to find the derivative. $$ f(x)=\frac{3}{\left(x^{3}-4\right)^{2}} $$
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Use a graphing utility to graph \(f\) and \(f^{\prime}\) on the interval \([-2,2]\). $$ f(x)=x^{2}(x+1)(x-1) $$
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Describe the interval(s) on which the function is continuous. \(f(x)=x \sqrt{x+3}\)
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Describe the \(x\) -values at which the function is differentiable. Explain your reasoning. $$ y=x^{2 / 5} $$
View solution