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Q53E

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Found in: Page 224

### Essential Calculus: Early Transcendentals

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.

See the step by step solution

## Step 1: Given data

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

## Step 2: Concept of Differentiation

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.

## Step 3: Substitute the value

(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)

## Step 4: Determine the value of the function

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 $$b = - 1$$.

## Step 5: Substitute the values

(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.