Question

How do you find a cubic function $y=a{{x}^{3}}+b{{x}^{2}}+cx+d$ whose graph has horizontal tangents at the points$\left[ \left( -2,6 \right)\,\,and\left( 2,0 \right) \right]?$

Answer 1

Horizontal tangents means that the slope of the line tangents to the cubic function is zero. Slope of the tangent line is the same as the derivative according to the first derivative test; the given points are local extremes.
First, set the derivatives of the cubic function equal to zero. Plug in both $x-$ values into the derivative. These are called critical points the location of the local extremes.
When we plug in the $x$values of points into the cubic function we get $y$values after getting equal we get values of $a,b,c\,\And d.$

Complete step by step solution:
The given function is
$f(x)=a{{x}^{3}}+b{{x}^{2}}+cx+d$
Hence,
$f\left( x \right)=a{{x}^{3}}+b{{x}^{2}}+cx+d$ the condition of horizontal tangency at points
\left\\{ {{x}_{1}},{{y}_{1}} \right\\},\left\\{ {{x}_{2}},{{y}_{2}} \right\\}is
Now differentiate function with respect to $x$
$\therefore \dfrac{df}{dx}f\left( x={{x}_{1}} \right)$
$\Rightarrow 3a{{x}_{1}}^{2}+2b{{x}_{1}}+c$
$\Rightarrow 0$
Now,
$\therefore \dfrac{df}{dx}f(x={{x}_{2}})$
$\Rightarrow 3a{{x}_{1}}^{2}+2b{{x}_{2}}+c$
$\Rightarrow 0$
Also we have, In horizontal tangency,
$\Rightarrow f\left( x={{x}_{1}} \right)=a{{x}_{1}}^{3}+b{{x}_{1}}^{2}+c{{x}_{1}}+d$
$={{y}_{1}}$
$\Rightarrow f\left( x={{x}_{2}} \right)=a{{x}_{2}}^{3}+b{{x}_{2}}^{2}+c{{x}_{2}}+d$
$={{y}_{2}}$
So we have the equation system. Write as follows
$\Rightarrow 12a-4b+c=0$
$\Rightarrow 12a+4b+c=0$
$\Rightarrow -8a+4b-2c+d=0$
$\Rightarrow 8a+4b+2c+d=0$
By solving this equations,
Solving the equation we get, value of $a,b,c\And d$ then after solving we get values of$a,b,c\And d$ are as follows.
$\therefore \,a=\dfrac{3}{16}$
$\therefore \,b=0$
$\therefore \,c=-\dfrac{9}{4}$
$\therefore \,d=3$

Note: First read and understand the question carefully. Always make sure that first step differentiate cubic equation that in first $\left( x={{x}_{1}} \right)\And$ and then $\left( x={{x}_{2.}} \right)$. Then after getting the equations. In the form of $a,b,c\And d.$
The values of $a,b,c\And d$ are found by solving equations. So make sure that during solving the equation, whatever you do to one side of the equation, whatever you do to one side of the equation, you have to do the same thing to the other side. Thus four steps to solving equations are as follows: addition, subtracting, multiplication & division. If you add the same number to both sides of an equation, both sides will remain equal.