Question

attempt 1: 2 attempts remaining. find the x-value(s) on the graph of f(x) = 2x³ + 12x² - 126x + 12 where the tangent is horizontal. if there is more than one x-value, enter them in a comma-separated list. if an x-value does not exist, enter dne. x =

Explanation:

Step1: Find the derivative of \( f(x) \)

The function is \( f(x) = 2x^3 + 12x^2 - 126x + 12 \). Using the power rule \( \frac{d}{dx}(x^n)=nx^{n - 1} \), the derivative \( f^\prime(x) \) is:
\( f^\prime(x)=2\times3x^{2}+12\times2x - 126\times1+0=6x^{2}+24x - 126 \)

Step2: Set the derivative equal to zero (horizontal tangent has slope 0)

We set \( f^\prime(x) = 0 \), so:
\( 6x^{2}+24x - 126 = 0 \)
Divide both sides of the equation by 6 to simplify:
\( x^{2}+4x - 21 = 0 \)

Step3: Solve the quadratic equation

For a quadratic equation \( ax^{2}+bx + c = 0 \) (here \( a = 1 \), \( b = 4 \), \( c=- 21 \)), we can use the quadratic formula \( x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \) or factor the equation. Let's factor:
We need two numbers that multiply to \( - 21 \) and add to \( 4 \). The numbers are \( 7 \) and \( - 3 \). So,
\( x^{2}+4x - 21=(x + 7)(x - 3)=0 \)
Setting each factor equal to zero gives:
\( x+7 = 0\) or \(x - 3=0 \)
Solving these, we get \( x=-7 \) or \( x = 3 \)

Answer:

\( -7, 3 \)