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Chapter 11: Problem 2

If $$ y=2 x^{3}+3 x^{2}-12 x+1 $$ find values of \(x\) for which \(y^{\prime \prime}=0\)

Short Answer

Expert verified
The value of \(x\) for which \(y'' = 0\) is \(-\frac{1}{2}\).

Step by step solution

01

Differentiate the function

Start by differentiating the function with respect to \(x\). Given \(y = 2x^3 + 3x^2 - 12x + 1\), the first derivative is obtained by applying the power rule. This gives: \[y' = \frac{d}{dx}(2x^3) + \frac{d}{dx}(3x^2) + \frac{d}{dx}(-12x) + \frac{d}{dx}(1) = 6x^2 + 6x - 12.\]
02

Differentiate the first derivative

To find the second derivative, \(y''\), differentiate the first derivative \(y' = 6x^2 + 6x - 12\). Apply the power rule again to each term: \[y'' = \frac{d}{dx}(6x^2) + \frac{d}{dx}(6x) + \frac{d}{dx}(-12) = 12x + 6.\]
03

Set the second derivative equal to zero

The problem asks us to find values of \(x\) for which the second derivative is zero. Therefore, set the expression for \(y''\) equal to zero and solve for \(x\): \[12x + 6 = 0.\]
04

Solve the linear equation

To solve \(12x + 6 = 0\), first subtract 6 from both sides: \[12x = -6.\]Then divide both sides by 12 to isolate \(x\): \[x = -\frac{6}{12} = -\frac{1}{2}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Derivatives
In calculus, a derivative represents how a function changes as its input changes. Think of it as the rate of change or the slope of the function at any particular point. For any function of the form \(y = f(x)\), the derivative \(y'\) (also written as \(\frac{dy}{dx}\)) provides the rate at which \(y\) changes with respect to \(x\).
  • When dealing with polynomials like \(y = 2x^3 + 3x^2 - 12x + 1\), finding the derivative will involve differentiating each term individually.
  • This effectively tells us how steep the curve of the function is at each point \(x\).
  • In practical terms, this could be akin to calculating speed when you want to know how fast something is moving at every moment.
The derivative is an essential tool in calculus for understanding behavior and trends within a function.
Exploring the Second Derivative
The second derivative, indicated as \(y''\) or \(\frac{d^2y}{dx^2}\), gives us information about the curvature or concavity of the function's graph.
  • While the first derivative \(y'\) relates to the slope, \(y''\) tells you how the slope itself is changing.
  • If \(y'' > 0\), the graph is concave up (shaped like a U), and if \(y'' < 0\), it is concave down (shaped like an upside-down U).
  • When \(y'' = 0\), it indicates a possible inflection point, where the curve changes from concave up to concave down, or vice versa.
In our problem, the task was solving \(12x + 6 = 0\) for \(x\), revealing the \(x\)-value at which the slope of the slope is neither increasing nor decreasing, thus indicating a change in curvature.
Applying the Power Rule
The power rule is a quick technique to find the derivative of polynomial functions.
  • The rule states: If \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\).
  • This means you bring the power (exponent) down front as a multiplier and decrease the power by one.
  • For example, consider \(2x^3\). Using the power rule, the derivative is \(3 \times 2x^{3-1} = 6x^2\).
Applying the power rule simplifies the differentiation process, turning potentially complex calculations into straightforward steps. It's a foundational tool in finding both first and second derivatives efficiently. Understanding this rule makes handling polynomial derivatives intuitive.

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Most popular questions from this chapter

Find \(\frac{\mathrm{d} y}{\mathrm{~d} x}\), given (a) \(x=t^{2} \quad y=1+t^{3}\) (b) \(x=\sin t \quad y=\mathrm{e}^{t}\) (c) \(x=(1+t)^{3} \quad y=1+t^{3}\)(d) \(x=\cos 2 t \quad y=3 t\) (e) \(x=\frac{3}{t} \quad y=\mathrm{e}^{2 t}\) (f) \(x=\mathrm{e}^{t}-\mathrm{e}^{-t} \quad y=\mathrm{e}^{t}+\mathrm{e}^{-t}\)

\text { Find } \frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}} \text { given } x y+x^{2}=y^{2} \text {. }

Differentiate each of the following functions: (a) \(y=5 \sin x\) (b) \(y=5 \mathrm{e}^{x} \sin x\) (c) \(y=5 \mathrm{e}^{\sin x}\) (d) \(y=\frac{5 \sin x}{\mathrm{e}^{-x}}\) (e) \(y=\left(t^{3}+4 t\right)^{15}\) (f) \(y=7 \mathrm{e}^{-3 t^{2}}\) (g) \(y=\frac{\sin x}{4 \cos x+1}\)

Calculate \(\frac{\mathrm{d} y}{\mathrm{~d} t}\) and \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} t^{2}}\) given (a) \(y=t^{2}+t\) (b) \(y=2 t^{3}-t^{2}+1\) (c) \(y=\sin 2 t\) (d) \(y=\sin k t \quad k\) constant (e) \(y=2 \mathrm{e}^{3 t}-t^{2}+1\) (f) \(y=\frac{t}{t+1}\) (g) \(y=4 \cos \frac{t}{2}\) (h) \(y=\mathrm{e}^{t} t\) (i) \(y=\sinh 4 t\) (j) \(y=\sin ^{2} t\)

Differentiate (a) \(y=\ln x\) (b) \(y=\ln 2 x\) (c) \(y=\ln k x, k\) constant (d) \(y=\ln (1+t)\) (e) \(y=\ln (3+4 t)\) (f) \(y=\ln (5+7 \sin x)\)
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