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Q53E

Expert-verifiedFound in: Page 224

Book edition
2nd

Author(s)
James Stewart

Pages
830 pages

ISBN
9781133112280

**(a) If the function \(f(x) = {x^3} + a{x^2} + bx\) has the local minimum value \( - \frac{2}{9}\sqrt 3 \) at \(x = \frac{1}{{\sqrt 3 }}\), what are the values of \(a\) and \(b\)?**

** **

**(b) Which of the tangent line of the curve in part (a) has the smallest slope?**

(a) The values of \(a = 0\) and \(b = - 1\).

(b) The tangent line \(x = 0\) has the smallest slope.

The given function is \(f(x) = {x^3} + a{x^2} + bx\).

**Differentiation is a method of finding the derivative of a function. Differentiation is a process, where we find the instantaneous rate of change in function based on one of its variables.**

(a)

Substitute the value as follows:

\(\begin{aligned}{c}f\left( {\frac{1}{{\sqrt 3 }}} \right) &= {\left( {\frac{1}{{\sqrt 3 }}} \right)^3} + a{\left( {\frac{1}{{\sqrt 3 }}} \right)^2} + b\left( {\frac{1}{{\sqrt 3 }}} \right)\\\frac{1}{{3\sqrt 3 }} + \frac{a}{3} + \frac{b}{{\sqrt 3 }} &= - \frac{2}{9}\sqrt 3 \\\frac{{1 + \sqrt 3 a + 3b}}{{3\sqrt 3 }} &= - \frac{2}{9}\sqrt 3 \\1 + \sqrt 3 a + 3b &= - 2\\\sqrt 3 a + 3b &= - 3\end{aligned}\)

If \(f\) has local minimum at \(x = \frac{1}{{\sqrt 3 }}\) then the derivative of the function at that point is zero That is \({f^\prime }\left( {\frac{1}{{\sqrt 3 }}} \right) = 0\)

Obtain derivative of \(f(x)\).

\(\begin{aligned}{l}{f^\prime }(x) &= \frac{d}{{dx}}\left( {{x^3} + a{x^2} + bx} \right)\\{f^\prime }(x) &= 3{x^2} + a(2x) + b\\{f^\prime }(x) &= 3{x^2} + 2ax + b\end{aligned}\) …….. (1)

Find the value of \({f^\prime }\left( {\frac{1}{{\sqrt 3 }}} \right) = 0\)

\(\begin{aligned}{c}{f^\prime }\left( {\frac{1}{{\sqrt 3 }}} \right) = 3{\left( {\frac{1}{{\sqrt 3 }}} \right)^2} + 2a\left( {\frac{1}{{\sqrt 3 }}} \right) + b = 0\\1 + \frac{{2a}}{{\sqrt 3 }} + b = 0\\\frac{{\sqrt 3 + 2a + b\sqrt 3 }}{{\sqrt 3 }} = 0\\\sqrt 3 + 2a + b\sqrt 3 = 0\\{\rm{ Thus, }}2a + \sqrt 3 b = - \sqrt 3 \end{aligned}\) ……… (2)

Solve the equations (1) and (2) and obtain \(a = 0\) and \(b = - 1\).

(b)

Substitute the values of \(a = 0\) and \(b = - 1\) in \(f(x)\),

\(\begin{aligned}{l}f(x) &= {x^3} + (0){x^2} + ( - 1)x\\f(x) &= {x^3} - x\end{aligned}\)

Obtain the derivative of \(f(x)\).

\(\begin{aligned}{c}{f^\prime }(x) &= \frac{d}{{dx}}\left( {{x^3} - x} \right)\\{f^\prime }(x) &= 3{x^2} - 1\end{aligned}\)

Thus, the slope of the tangent line to the curve is, \({f^\prime }(x) = m(x) = 3{x^2} - 1\).

Obtain the derivative of the slope.

\(\begin{aligned}{l}{m^\prime }(x) &= \frac{d}{{dx}}\left( {3{x^2} - 1} \right)\\{m^\prime }(x) &= 3(2x) - 0\\{m^\prime }(x) &= 6x\end{aligned}\)

To obtain the smallest slope set \({m^\prime }(x) = 0\).

\(\begin{aligned}{c}6x = 0\\x = 0\end{aligned}\)

The second derivative of the slope is, \({m^{\prime \prime }}(x) = 6\), which is greater than 0.

So, the tangent line to the curve which occurs at \(x = 0\) has the minimum slope.

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