Question

Find the points on the curve \(y = {x^3} + 3{x^2} - 9x + 10\) where the tangent is horizontal.

Short answer

The points at which the curve is horizontal are \(\left( { - 3,37} \right)\)and \(\left( {1,5} \right)\).

Step-by-step solution

Differentiate the given function

Differentiate the given function with respect to\(x\), as shown follows:

\(\begin{aligned}y' &= \frac{d}{{dx}}\left( {{x^3} + 3{x^2} - 9x + 10} \right)\\ &= \frac{d}{{dx}}\left( {{x^3}} \right) + \frac{d}{{dx}}\left( {3{x^2}} \right) - \frac{d}{{dx}}\left( {9x} \right) + \frac{d}{{dx}}\left( {10} \right)\\ &= 3{x^2} + 3\left( {2x} \right) - 9\left( 1 \right) + 0\\ &= 3{x^2} + 6x - 9\end{aligned}\)

Therefore, \(y' = 3{x^2} + 6x - 9\).

Apply the condition of Horizontal tangent

The horizontal tangents occur at those values of\(x\)at which\(y' = 0\). So, set the obtained expression of\(y'\)equal to 0 and solve for\(x\)as follows:

\(\begin{aligned}3{x^2} + 6x - 9 &= 0\\3\left( {{x^2} + 2x - 3} \right) &= 0\\3\left( {x + 3} \right)\left( {x - 1} \right) &= 0\\x &= - 3,\,\,1\end{aligned}\).

Find the value of function at \(x =  - 3,1\)

Finally, solve the givenfunction at\(x = - 3,\,1\)by plugging\(x = - 3,\,1\)into the expression for\(y\)as shown below:

\(\begin{aligned}y &= {\left( { - 3} \right)^2} + 3{\left( { - 3} \right)^2} - 9\left( { - 3} \right) + 10\\ &= 37\end{aligned}\)

\(\begin{aligned}y &= {\left( 1 \right)^2} + 3{\left( 1 \right)^2} - 9\left( 1 \right) + 10\\ &= 5\end{aligned}\)

Thus, the points at which the curve is horizontal are \(\left( { - 3,37} \right)\)and \(\left( {1,5} \right)\).