Question

find all the value(s) of x at which (f(x)=\frac{x^{3}}{1}+4x^{2}+16x + 1) has a horizontal tangent line. if there is more than one answer, give all of the x - values separated by commas, e.g. if (f(x)) has a horizontal tangent line at (x = 3) and (x = 5) enter 3, 5

Explanation:

Step1: Find the derivative

The derivative of $f(x)=x^{3}+4x^{2}+16x + 1$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ is $f^\prime(x)=3x^{2}+8x + 16$.

Step2: Set the derivative equal to 0

A horizontal tangent line occurs when $f^\prime(x)=0$. So we solve the quadratic equation $3x^{2}+8x + 16 = 0$. The quadratic formula for a quadratic equation $ax^{2}+bx + c=0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 3$, $b = 8$, and $c = 16$.

Step3: Calculate the discriminant

The discriminant $\Delta=b^{2}-4ac=(8)^{2}-4\times3\times16=64 - 192=-128<0$. Since the discriminant is negative, the quadratic equation $3x^{2}+8x + 16 = 0$ has no real solutions.

Answer:

No real - valued solutions