Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration

Q53E

Expert-verified
Essential Calculus: Early Transcendentals
Found in: Page 224
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

Answers without the blur.

Just sign up for free and you're in.

Illustration

Short Answer

(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 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.

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.