Skip to content

Lars Eighner's Homepage


LarsWiki

calumus meretrix et gladio innocentis

Calculus Polynomial Graph Exercises Worked


Exercises from public domain textbooks. Solutions where provided do not necessary follow the same lines as the original.

Contents

D33-1. {$$ y = x^3 - 6x^2 + 9x + 3 $$}

Inspection makes clear {$y=3, \text{ when } x=0 $}. X-intercepts, if any, are not so clear. It should also be clear that for large values of x, y increases without bound.

The first derivative, by inspection, factors easily:

{$$ \begin{align} y^\prime &= 3x^2 - 12x + 9 \cr &= 3(x-1)(x-3) \cr \text{Critical points: } x&=1 \cr x&=3 \cr y^{\prime\prime} &= 6x - 12 \cr &= 6(x-2) \end{align} $$}

The second derivative is negative on {$ (-\infty,2) $}, positive on {$ (2,\infty) $}, and changes sign at {$ x= 2$}. As the critical point at {$ x = 1 $} is in the negative part of the second derivative, we conclude that it is a maxima, and by the same token, the point at {$ x=2 $} is a minima.

They values for these points are found by plugging into the original equation, which is simple arithmetic and is not shown.

The slope of the tangent at the inflection point is found by substituting in the first derivative, and the equation of the tangent at the inflection point is found from the point-slope formula for a line.

{$$ \begin{align} m &= y^\prime = 3x^2 - 12x + 9 \cr m_2 &= 3(2)^2 - 12(2) + 9 \cr m_2 &= -3 \cr y-y_0 &= m(x-x_0) \tag{slope-intercept formula} \cr y-5 &= (-3)(x - 2) \cr y&= -3x -6 +5 \cr y&= -3x - 1 \end{align} $$}

Points
xyTypeMethod
03interceptinspection
17maximumfirst derivative 0
33minimumfirst derivative 0
25inflectionsecond derivative 0

{$$ \text{D33-1. } \quad y = x^3 - 6x^2 + 9x + 3 $$}

D33-2. {$$ y = 4 + 3x - x^3 $$}

By inspection it is clear that the y-intercept is at 4, and that as x gets large, the negative cubed term will outstrip the others, and y will decrease without bound.

{$$ \begin{align} y &= 4 + 3x - x^3 \cr y^\prime &= -3x^2 + 3 \cr &=3(1- x^2) = 3(x+1)(x-1) \cr y^{\prime\prime} &= -6x \end{align} $$}

So the first derivative has zeros at {$x=1$} and {$x=-1$}, and the second derivative is negative when x is positive and positive where x is negative, with the inflection point being at {$(0,4)$}, which happens to be where the y-intercept is.

The function is concave upward when x is negative (the second derivative is positive), so the critical point at {$(-1,2)$} is a minima, and the critical point at {$(1,6)$} is a maxima as it in the region where the function is concave downwards (second derivative is negative).

The tangent at the inflection point is found by plugging into the first derivative to get the slope, and then using the point-slope formula to find the equation of the line.

{$$ \begin{align} m &= f^\prime(x) = -3x^2 + 3 \cr m_0 &= f^\prime(0) = 3(0)^2 + 3 \cr m_0 &= 3 \cr y-y_0 &= m(x-x_0) \tag{slope-intercept formula} \cr y-4 &= (-3)(x - 0) \cr y&= 3x + 4 \end{align} $$}

Points
xyTypeMethod
04interceptinspection
16maximumfirst derivative 0
-12minimumfirst derivative 0
04inflectionsecond derivative 0

{$$ \text{D33-2. } \quad y = 4 + 3x - x^3 $$}

D33-3. {$$ y = x^3 - 3x^2 + 6x +10 $$}

By inspection, the y-intercept is at {$(0,10)$}. Also, the coefficients and constant seem suspicious, and a little thought reveals an x-intercept at {$(-1,0)$}. As x gets large, y rapidly increases without bound.

{$$ \begin{align} y &= x^3 - 3x^2 + 6x +10 \cr y^\prime &= 3x^2 -6x + 6 \cr y^{\prime\prime} &= 6x-6 = 6(x-1) \end{align} $$}

The first derivative has no real roots, and it is easy to see that it is always positive. So there are no critical points. The second derivative reveals there is a point of inflection at {$x=1$}.

The tangent at the inflection point is found by plugging into the first derivative to get the slope, and then using the point-slope formula to find the equation of the line.

{$$ \begin{align} m &= f^\prime(x) = 3x^2 -6x +6 \cr m_1 &= f^\prime(1) = 3(1)^2 - 6(1) + 6 \cr m_1 &= 3 \cr y-y_0 &= m(x-x_0) \tag{slope-intercept formula} \cr y-14 &= (3)(x - 1) \cr y&= 3x + 11 \end{align} $$}

Points
xyTypeMethod
010interceptinspection
-10interceptinspection
114inflectionsecond derivative 0

{$$ \text{D33-3. } \quad y = x^3 - 3x^2 + 6x +10 $$}

D33-4. {$$ y = (x-3)^2(x-2) $$}

This function is conveniently presented in factored form. So, it is clear from inspection that the x-intercepts are at {$(2,0)$} and {$(3,0)$}. Moreover, it is easy to see the constant term, so the y-intercept is at {$(0,-18)$}.

This form is convenient for many calculations, but it is a wash whether to differentiate it by the product rule or to multiply it out and then differentiate.

{$$ \begin{align} y &= (x-3)^2(x-2) \cr y &= x^3 - 8x^2 + 21x -18 \cr y^\prime &= 3x^2 - 16x + 21 \cr y^\prime &= (3x - 7)(x - 3) \cr y^{\prime\prime} &= 6x - 16 \end{align} $$}

The critical points are the zeros of the first derivative which are at {$ x = {7 \over 3} $} and {$ x = 3 $}. The second derivative is negative for {$ x \lt {8 \over 3} $}, so {$ ({7 \over 3},{4 \over 27})$} is a maximum in this region that is concave downward. The second derivative is positive for {$ x \gt {8 \over 3} $} so {$ (3,0) $} is a minimum in the concave upward region. The second derivative changes sign at {$ ({8 \over 3}, {2 \over 27} ) $}, so this is a point of inflection.

The slope of the tangent at the inflection point is found by substituting in the first derivative, and the equation of the tangent at the inflection point is found from the point-slope formula for a line.

{$$ \begin{align} m &= 3x^2 - 16x + 21 \cr m_{8 / 3} &= 3\left( {8 \over 3} \right) ^2 - 16\left( {8 \over 3} \right) +21 \cr &= 3 \left( {64 \over 9} \right) - \left( {128 \over 3} \right) +21 \cr &= {64 \over 3} - {128 \over 3} + 21 \cr m_{8 / 3} &= -21{1 \over 3} + 21 = -{1 \over 3} \cr y-y_0 &= m(x-x_0) \tag{slope-intercept formula} \cr y-{2 \over 27} &= -{1 \over 3} \left( x - {8 \over 3} \right) \cr y &= -{1 \over 3}x - \left( {8 \over 3} \right) \left( -{1 \over 3} \right) + {2 \over 27} \cr y &= -{1 \over 3}x + {26 \over 27} \end{align} $$}

Points
xyTypeMethod
20interceptinspection
30interceptinspection
0-18interceptinspection
7/34/27maximumfirst derivative 0
30minimumfirst derivative 0
8/32/27inflectionsecond derivative 0

{$$ \text{D33-4. } \quad y = (x-3)^2(x-2) $$}

D33-5. {$$ y = (1-x^2)^3 $$}

Warning! This working does not agree with the textbook answer.

The intercepts, evident on inspection, turn out to be critical points.

{$$ \begin{align} y &= (1-x^2)^3 \cr {d \over {dx}} y &= 3(1-x^2)^2(-2x) \tag{Chain rule} \cr &= -6x(1-x^2)^2 \end{align} $$}

This make the critical points at the zeros of the first derivative {$(0,1), (1,0), \text{ and } (-1,0) $}. Things get hairy with the second derivative, which in consideration of the difference between this solution and the textbook answer suggest close scrutiny.

{$$ \begin{align} {d \over {dx}} y &= -6x(1-x^2)^2 \cr {d \over {d^2x}}y &= (1-x^2)^2{d \over {dx}}(-6x) + (-6x){d \over {dx}}\left[ (1-x^2)^2 \right] \tag{Product rule} \cr &= -6(1-x^2)^2 + (-6x)2(1-x^2)(-2x) \tag{Chain rule} \cr &= -6 + 12x^2 - 6x^4 + 24x^2 - 24x^4 \cr &= -30x^4 + 36x^2 -6 \cr {d \over {d^2x}}y &= (6 - 30x^2)(1-x^2) \end{align} $$}

A pair of roots here is undisputed {$ x = \pm 1$}, but it seems to me the other pair is {$ x = \pm {1 \over 5}\sqrt{5} $} (not the textbook answer of {$ x = \pm {1 \over 3}\sqrt{3} $}). I have worked this a dozen times with varying degrees of hairy algebra, and I just do not see my error.

As this is an even function, it does not matter which root we use to find the y values:

{$$ \begin{align} f(x) &= (1-x^2)^2 \cr f({{\sqrt{5}} \over 5}) &= \left[ 1 - \left( {{\sqrt{5}} \over 5} \right)^2 \right] ^2 \cr &= ({4 \over 5})^2 = {16 \over 25} \end{align} $$}

To take a closer look at the factors of the second derivative: {$ (6-30x^2)(x^2 - 1)$}, clearly when {$ x \gt 1 \text{ or } x \lt 1 $}, {$x^2 - 1$} is positive, {$6-30x^2$} is negative, so the sign of the second derivative is negative, but for {$ -1 \lt x \lt 1, x^2 - 1 \lt 0 $}, so the sign of the second derivative is simple the opposite of whatever the sign of {$ 6 - 30x^2 $} is.

Range{$ x^2 -1$}{$ 6-30x^2 $}{$ y^{\prime\prime} $}Concave
{$$ x \lt -1 $$}positivenegativenegativedownward
{$$ -1 \lt x \lt -{\sqrt{5} \over 5} $$}negativenegativepositiveupward
{$$ -{\sqrt{5} \over 5} \lt x \lt {\sqrt{5} \over 5} $$}negativepositivenegativedownward
{$$ {\sqrt{5} \over 5} \lt x \lt 1 $$}negativenegativepositiveupward
{$$ 1 \lt x $$}positivenegativenegativedownward

The four zeros of the second derivative divide x into five regions, and the sign of the second derivation changes at each of the points, so all four are points of inflection. The points of inflection account for two of the critical points, found from the zeros of the first derivative. The third critical point falls in the middle of a concave downward region and is, thereby, a maximum.

Clearly the tangent lines at {$ x = \pm 1 $} are one and the same, namely the x-axis, with equation {$ y = 0 $}. To find the tangents at the other points of inflection, the second derivative is consulted to determine their slopes:

{$$ \begin{align} m &= f'[\pm (5)^{-{1 \over 2}}] \cr &= -6[\pm (5)^{-{1 \over 2}}]\left(1 - [\pm (5)^{-{1 \over 2}}]^2 \right) ^2 \cr &= \mp 6(5)^{-{1 \over 2}}\left(1 - (5)^{-1} \right) ^2 \cr &= \mp 6(5)^{-{1 \over 2}}\left(4(5)^{-1} \right) ^2 \cr &= \mp 6(5)^{-{1 \over 2}}(16)(5)^{-2} \cr &= \mp 96(5)^{-{5 \over 2}} \end{align} $$}

Notice for the negative root, the slope is positive, and vice versa. Then into the point slope formula, to find the lines:

{$$ \begin{align} y-y_0 &= m(x-x_0) \tag{point-slope formula} \cr y - 64(5)^{-3} &= \pm 96(5)^{-{5 \over 2}}\left( x - (\mp (5)^{-{1 \over 2}} )\right) \end{align} $$}

Notice the constant term must always come out with a positive sign.

{$$ \begin{align} y - 64(5)^{-3} &= \pm 96(5)^{-{5 \over 2}}\left( x - (\mp (5)^{-{1 \over 2}} )\right) \cr y &= \pm 96(5)^{-{5 \over 2}}( x ) + 96(5)^{-{5 \over 2}}(5)^{-{1 \over 2}} ) + 64(5)^{-3} \cr y &= \pm 96(5)^{-{5 \over 2}}( x ) + 96(5)^{-3} + 64(5)^{-3} \cr y &= \pm 96(5)^{-{5 \over 2}}( x ) + 160(5)^{-3} \cr y &= \pm 96(5)^{-{5 \over 2}}( x ) + 1.28 \cr y &\approx \pm 1.717( x ) + 1.28 \cr \end{align} $$}

Points
xyTypeMethod
10interceptinspection
-10interceptinspection
01interceptinspection
01maximumfirst derivative 0
10criticalfirst derivative 0
-10criticalfirst derivative 0
10inflectionsecond derivative 0
-10inflectionsecond derivative 0
√5/564/125inflectionsecond derivative 0
-√5/564/125inflectionsecond derivative 0

{$$ \text{D33-5. } \quad y = (1-x^2)^3 $$}

D33-6. {$$ y = (4-x^2)^2 $$}

Since this whole function is a square, it will be everywhere positive and symmetric about the y-axis. For large values of x, either positive or negative, it will increase rapidly without bound. As it is the square of the difference of two squares, the roots are easily extracted by inspection.

{$$ \begin{align} y &= (4-x^2)^2 \cr y^\prime &= -4x(4-x^2) \cr &= 4x^3 - 16x \cr y^{\prime\prime} &= 12x^2 - 16 \end{align} $$}

Range{$ y^{\prime\prime} $}Concave
{$$ x \lt -{2 \over 3}\sqrt{3} $$}positiveupward
{$$ -{2 \over 3}\sqrt{3} \lt x \lt {2 \over 3}\sqrt{3} $$}negativedownward
{$$ x \gt {2 \over 3}\sqrt{3} $$}positiveupward

To obtain the slope at the inflection points, plug into the second derivative:

{$$ \begin{align} m = f^\prime(\pm {2 \over 3}\sqrt3 ) &= -4(\pm {2 \over 3}\sqrt3 )(4 - (\pm {2 \over 3}\sqrt3 )^2) \cr &= \mp {8 \over 3}\sqrt3 (4 - ({{4 \cdot 3} \over 9})) \cr &= \mp {8 \over 3}\sqrt3 ({24 \over 9}) \cr m &= \mp {64 \over 9}\sqrt3 \end{align} $$}

And then to plug into the point-slope formula to find the tangent lines at the intercepts:

{$$ \begin{align} y - {64 \over 9} &= \mp {64 \over 9}\sqrt3 \left( x - (\pm {2 \over 3}\sqrt3 \right) \cr y - {64 \over 9} &= \mp {64 \over 9}\sqrt3 \left( x \right) - (\mp {64 \over 9}\sqrt3)(\pm {2 \over 3}\sqrt3 ) \cr y &= \mp {64 \over 9}\sqrt3 \left( x \right) + {128 \over 9} + {64 \over 9} \cr y &= \mp {64 \over 9}\sqrt3 \left( x \right) + {64 \over 3} \end{align} $$}

Points
xyTypeMethod
20interceptinspection
-20interceptinspection
016interceptinspection
016maximumfirst derivative 0
20minimumfirst derivative 0
-20minimumfirst derivative 0
(2/3)√364/9inflectionsecond derivative 0
-(2/3)√364/9inflectionsecond derivative 0

{$$ \text{D33-6. } \quad y = (4-x^2)^2 $$}

D33-7. {$$ y = (x-1)^3(x+2)^2 $$}

The intercepts are easily found by inspection. As x gets large, this function increases without bound. As x gets increasing negative, the function decreases without bound.

{$$ \begin{align} y &= (x-1)^3(x+2)^2 \cr y^\prime &= (x+2)^2(3)(x-1)^2 + (x-1)^3(2)(x+2) \end{align} $$}

Not shown are the Chain rule results because the derivatives of the interior functions are simply a factor of 1.

{$$ \begin{align} y^\prime &= (x+2)^2(3)(x-1)^2 + (x-1)^3(2)(x+2) \cr &= (x-1)^2(x+2) \left[ (x+2)(3) + (x-1)(2) \right] \cr y^\prime &= (x-1)^2(x+2)(5x+4) \end{align} $$}

To find the new root and its y value: {$$ \begin{align} 5x +4 &= 0 \cr x &= - {4 \over 5} \cr f(- {4 \over 5}) &= (-{4\over5} -1)^3(-{4\over5} + 2)^2 \cr &= (-{9\over5})^3({6\over5})^2 = -{2^2}{3^8}5^{-5} \approx -8.4 \end{align} $$}

Taking the second derivative will involve hairy algebra whether done by the Product rule or by expanding the polynomial first.

{$$ \begin{align} y^\prime &= (x-1)^2(x+2)(5x+4) \cr {d \over {dx}} y^\prime &= {d \over {dx}}\left[ (x-1)^2(x+2)(5x+4) \right] \cr &= (5x+4){d \over {dx}}\left[ (x-1)^2(x+2) \right] + (x-1)^2(x+2){d \over {dx}}(5x+4) \cr &= (5x+4)\left[ (x+2){d \over {dx}}(x-1)^2 +(x-1)^2{d \over {dx}}(x+2) \right] \cr &+ (x-1)^2(x+2)(5) \cr &= (5x+4)(x+2)(2)(x-1) +(5x+4)(x-1)^2(1) \cr &+ (x-1)^2(5x+10) \cr &= (x-1) \left[ (5x+4)(x+2)(2) +(5x+4)(x-1) + (x-1)(5x+10) \right] \cr &= (x-1) \left[ 10x^2+28x+16 + 5x^2 -x -4 + 5x^2+5x -10 \right] \cr &= (x-1) ( 20x^2+32x+2) \end{align} $$}

Unfortunately, the latter quadratic does not factor, so it is necessary to resort to the quadratic formula to find the other two roots.

{$$ \begin{align} x &= \frac{-32 \pm \sqrt{\left[ (32)^2 - 4(20)(2) \right] }}{2(20)} \cr x &= -{4 \over 5} \pm {{3\sqrt{6}} \over 10} \end{align} $$}

Range{$ y^{\prime\prime} $}Concave
{$$ -{4 \over 5} - {{3\sqrt{6}} \over 10} \gt x $$}negativedownward
{$$ -{4 \over 5} - {{3\sqrt{6}} \over 10} \lt x \lt -{4 \over 5} + {{3\sqrt{6}} \over 10} $$}positiveupward
{$$ -{4 \over 5} + {{3\sqrt{6}} \over 10} \lt x \lt 1 $$}negativedownward
{$$ 1 \lt x $$}positiveupward

The tangent at {$x=1$} is the x-axis which has slope 0. to find the slopes of the tangents at the other two inflection points, plug into the first derivative. This seems to be slightly easier with the factored form:

{$$ \begin{align} y^\prime &= (x-1)^2(x+2)(5x+4) \cr m &= \left[-4/5 \pm {{3\sqrt{6}} \over 10} -1 \right]^2 \left[ -4/5 \pm {{3\sqrt{6}} \over 10} +2 \right] \left[ 5 \left( -4/5 \pm {{3\sqrt{6}} \over 10} \right) +4) \right] \cr &= \left[ \pm {{3\sqrt{6}} \over 10} - {9 \over 5} \right]^2 \left[ \pm {{3\sqrt{6}} \over 10} + {5 \over 6} \right] \left[ \pm {{3\sqrt{6}} \over 2} \right] \cr &= \left[ \mp {{27\sqrt{6}} \over 25} + {189 \over 50} \right] \left[ \pm {{3\sqrt{6}} \over 10} + {5 \over 6} \right] \left[ \pm {{3\sqrt{6}} \over 2} \right] \cr &= \left[ \mp {{27\sqrt{6}} \over 25} + {189 \over 50} \right] \left[ \pm {{9\sqrt{6}} \over 5} + {27 \over 10} \right] \tag{1st factor is positive} \cr m &= \pm {{486\sqrt{6}} \over 125} - {729 \over 500} \end{align} $$}

And into the point slope formula to find the equations of the tangent lines at the inflection points (other than {$ x=1 $}.

{$$ \begin{align} y-y_0 &= m(x-x_0) \cr y- \left(\mp {5103 \over 25000}\sqrt{6} - {50301 \over 12500} \right) &= \left( \pm {{486\sqrt{6}} \over 125} - {729 \over 500} \right)\left[ x-\left( -{4 \over 5} \pm {3 \over 10}\sqrt{6} \right) \right] \cr \end{align} $$}

{$$ \begin{align} y &= \left( \pm {{486\sqrt{6}} \over 125} - {729 \over 500} \right)\left[ x-\left( -{4 \over 5} \pm {3 \over 10}\sqrt{6} \right) \right]+ \left(\mp {5103 \over 25000}\sqrt{6} - {50301 \over 12500} \right) \cr &= \left( \pm {{486\sqrt{6}} \over 125} - {729 \over 500} \right) (x)- \left( \pm {{486\sqrt{6}} \over 125} - {729 \over 500} \right) \left( -{4 \over 5} \pm {3 \over 10}\sqrt{6} \right) \cr &+ \left(\mp {5103 \over 25000}\sqrt{6} - {50301 \over 12500} \right) \cr y &= \left( \pm {{486\sqrt{6}} \over 125} - {729 \over 500} \right) (x) \mp {{10449\sqrt{6}} \over 3125} - {152361 \over 12500} \end{align} $$}

Points
xyTypeMethod
10interceptinspection
20interceptinspection
0-4interceptinspection
10criticalfirst derivative 0
20maximumfirst derivative 0
-4/5{$ -{26244 \over 3125} \approx -8.4 $}minimumfirst derivative 0
10inflectionsecond derivative 0
{$$ -{4 \over 5} - {3 \over 10}\sqrt{6} $$}{$$ {5103\over 25000}\sqrt{6} - {50301 \over 12500} \approx -3.524 $$}inflectionsecond derivative 0
{$$ -{4 \over 5} + {3 \over 10}\sqrt{6} $$}{$$ - {5103 \over 25000}\sqrt{6} - {50301 \over 12500} \approx -4.524 $$}inflectionsecond derivative 0

In checking against textbook answers, note that {$\sqrt{54} = 3\sqrt{6} $}.


{$$ \text{D33-7. } \quad y = (x-1)^3(x+2)^2 $$}


Sources:

  1. Elements of the Differential and Integral Calculus Wikisource
  2. Calculus Made Easy by Silvanus P. Thompson Project Gutenberg
  3. Davis,Ellery Williams, William Charles Brenke, Earle Raymond Hedrick. The Calculus (Macmillan Company, 1922) Google Books
  4. Love, Clyde E., Earl David Rainville Differential and Integral Calculus (Macmillian, 1916) Google Books

Recommended:

Category: Math Calculus


Read or Post Comments

No comments yet.

This is a student's notebook. I am not responsible if you copy it for homework, and it turns out to be wrong.

Figures are often enhanced by hand editing; the same results may not be achieved with source sites and source apps.

Backlinks

This page is MathAidenExercisesWorked

August 05, 2017

  • HomePage
  • WikiSandbox

Lars

Contact by Snail!

Lars Eighner
APT 1191
8800 N IH 35
AUSTIN TX 78753
USA

Help

HOME

The best way to look for anything in LarsWiki is to use the search bar.

Page List

Categories

Physics Pages

Math Pages

Math Exercises

Math Tools

Sections