Question

determine the critical values of ( f(x) = 4x^3 + 3x^2 - 6x + 14 ). separate your answers with commas. leave answers as integers or reduced fractions.
the critical values are:

Explanation:

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

The function is \( f(x) = 4x^3 + 3x^2 - 6x + 14 \). Using the power rule \( \frac{d}{dx}(x^n)=nx^{n - 1} \), the derivative \( f'(x) \) is:
\( f'(x)=4\times3x^{2}+3\times2x-6\times1 + 0=12x^{2}+6x - 6 \)

Step2: Set the derivative equal to zero

To find critical values, we solve \( f'(x) = 0 \), so:
\( 12x^{2}+6x - 6 = 0 \)
We can divide the entire equation by 6 to simplify:
\( 2x^{2}+x - 1 = 0 \)

Step3: Solve the quadratic equation

We factor the quadratic equation \( 2x^{2}+x - 1 = 0 \). We need two numbers that multiply to \( 2\times(- 1)=-2 \) and add up to 1. The numbers are 2 and -1. So we rewrite the middle term:
\( 2x^{2}+2x - x - 1 = 0 \)
Factor by grouping:
\( 2x(x + 1)-1(x + 1)=0 \)
\( (2x - 1)(x + 1)=0 \)
Setting each factor equal to zero gives:
\( 2x - 1 = 0 \) or \( x + 1 = 0 \)
For \( 2x - 1 = 0 \), we have \( 2x=1 \), so \( x=\frac{1}{2} \)
For \( x + 1 = 0 \), we have \( x=-1 \)

Answer:

\( -1,\frac{1}{2} \)