Question

How do you find the points where the graph of the function \[f\left( x \right) = 3{x^3} + 8{x^2} + 4x + 7\] has horizontal tangents and what is the equation?

Answer 1

Hint: This question involves the arithmetic operations like addition/ subtraction/ multiplication/ division. We need to know the basic form of a quadratic equation and the formula to find the value \[x\] from the quadratic equation. Also, we need to know the process of finding the slope of the tangents from the given equation.

Complete step by step solution:
The given equation is shown below,
 \[f\left( x \right) = 3{x^3} + 8{x^2} + 4x + 7 \to \left( 1 \right)\]
To find the slope of the tangent we have to find the derivation of the above equation. So, we get
 \[{f^1}\left( x \right) = \dfrac{{df\left( x \right)}}{{dx}}\]
 \[{f^1}\left( x \right) = \dfrac{d}{{dx}}\left( {3{x^3} + 8{x^2} + 4x + 7} \right) \to \left( 2 \right)\]
We know that,
 \[\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}\]
By using this formula the equation \[\left( 2 \right)\] , becomes,
 \[\left( 2 \right) \to {f^1}\left( x \right) = \dfrac{d}{{dx}}\left( {3{x^3} + 8{x^2} + 4x + 7} \right)\]
 \[{f^1}\left( x \right) = \left( {3 \times 3{x^2}} \right) + \left( {8 \times 2x} \right) + 4 + 0\]
 \[{f^1}\left( x \right) = 9{x^2} + 16x + 4\]
We know that the tangent is horizontal when the slope is zero. So, we get
 \[
  {f^1}\left( x \right) = 0 \\
  {f^1}\left( x \right) = 9{x^2} + 16x + 4 = 0 \to \left( 3 \right) \\
 \]
Let’s solve the above equation by using the quadratic formula,
We know that
If \[a{x^2} + bx + c = 0\] , then
 \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] \[ \to \left( 4 \right)\]
From the equation \[\left( 3 \right)\] , we get
 \[a = 9,b = 16,c = 4\]
So, the equation \[\left( 4 \right)\] becomes,
 \[
  \left( 4 \right) \to x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
  x = \dfrac{{ - 16 \pm \sqrt {{{\left( {16} \right)}^2} - 4 \times 9 \times 4} }}{{2 \times 9}} = \dfrac{{ - 16 \pm \sqrt {256 - 144} }}{{18}} = \dfrac{{ - 16 \pm \sqrt {112} }}{{18}} \\
 \]
So, we get
 \[x = \dfrac{{ - 16 \pm \sqrt {4 \times 4 \times 7} }}{{18}} = \dfrac{{ - 16 \pm 4\sqrt 7 }}{{18}} = \dfrac{{2\left( { - 8 \pm 2\sqrt 7 } \right)}}{{18}}\]
By solving the above equation we get,
 \[x = \dfrac{{ - 8 \pm 2\sqrt 7 }}{9}\]
Let’s solve the above equation using two cases,
Case: \[1\]
 \[x = \dfrac{{ - 8 + 2\sqrt 7 }}{9} = \dfrac{{ - 8 + 5.291}}{9} = - 0.301\]
Case: \[2\]
 \[x = \dfrac{{ - 8 - 2\sqrt 7 }}{9} = \dfrac{{ - 8 - 5.291}}{9} = - 1.476\]
So, we get
 \[x = - 0.301, - 1.476\]
Let’s substitute these values in the equation \[\left( 1 \right)\] ,
Take \[x = - 0.301\]
 \[\left( 1 \right) \to f\left( x \right) = 3{x^3} + 8{x^2} + 4x + 7\]
 \[f\left( x \right) = 3{\left( { - 0.301} \right)^3} + 8{\left( { - 0.301} \right)^2} + 4\left( { - 0.301} \right) + 7\]
 \[
  f\left( x \right) = - 0.081 + 0.724 - 1.204 + 7 \\
  f\left( x \right) = 6.439 \;
 \]
Take \[x = - 1.476\]
 \[\left( 1 \right) \to f\left( x \right) = 3{x^3} + 8{x^2} + 4x + 7\]
 \[
  f\left( x \right) = 3{\left( { - 1.476} \right)^3} + 8{\left( { - 1.476} \right)^2} + 4\left( { - 1.476} \right) + 7 \\
  f\left( x \right) = - 9,646 + 17.428 - 5.904 + 7 \;
 \]
 \[f\left( x \right) = 8.878\]
So, the final answer is,
Horizontal tangent equations:
 \[f\left( x \right) = 6.439\] At \[\left( { - 0.301,6.439} \right)\]
 \[f\left( x \right) = 8.878\] At \[\left( { - 1.476,8.878} \right)\]
So, the correct answer is “ \[f\left( x \right) = 6.439\] At \[\left( { - 0.301,6.439} \right)\]
 \[f\left( x \right) = 8.878\] At \[\left( { - 1.476,8.878} \right)\] ”.

Note: Note that the denominator term would not be equal to zero. Remember the quadratic formula to solve quadratic equations. Note that when the slope is zero, the tangent is horizontal. Note that when we differentiate the constant term the answer will be zero. It \[ \pm \] is present in one term we would find two answers for the term.