For the following function, find (a) the critical numbers, (b) the open intervals where the function is increasing, and (c) the open intervals where it is decreasing. $\newline$ $f(x)=x^{4}+12 x^{3}+28 x^{2}+4$

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Q. For the following function, find (a) the critical numbers, (b) the open intervals where the function is increasing, and (c) the open intervals where it is decreasing.

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f(x)=x^{4}+12 x^{3}+28 x^{2}+4

Find Derivative of $f(x)$ : First, find the derivative of $f(x)$ to locate critical numbers. $f'(x) = 4x^{3} + 36x^{2} + 56x + 0$
Locate Critical Numbers: Set the derivative equal to zero to find critical points. $4x^{3} + 36x^{2} + 56x = 0$
Set Derivative Equal to Zero: Factor out the greatest common factor, which is $4x$ . $\newline$ $4x(x^{2} + 9x + 14) = 0$
Factor Out Common Factor: Now factor the quadratic equation. $\newline$ $x(x + 7)(x + 2) = 0$
Factor Quadratic Equation: Solve for $x$ to find the critical numbers. $x = 0$ , $x = -7$ , $x = -2$
Solve for Critical Numbers: Use a sign chart or test values in the intervals to determine where $f'(x)$ is positive or negative. $\newline$ Test intervals: $(-\infty, -7)$ , $(-7, -2)$ , $(-2, 0)$ , $(0, \infty)$
Determine Sign of $f'(x)$ : Choose test points: $-8$ , $-5$ , $-1$ , $1$ and plug into $f'(x)$ . $f'(-8) = 4(-8)^{3} + 36(-8)^{2} + 56(-8) > 0$ $f'(-5) = 4(-5)^{3} + 36(-5)^{2} + 56(-5) < 0$ $f'(-1) = 4(-1)^{3} + 36(-1)^{2} + 56(-1) > 0$ $f'(1) = 4(1)^{3} + 36(1)^{2} + 56(1) > 0$
Test Intervals with Points: Determine the intervals of increase and decrease. $\newline$ $f'(x) > 0$ on $(-\infty, -7)$ and $(-2, 0)$ and $(0, \infty)$ so $f(x)$ is increasing on these intervals. $\newline$ $f'(x) < 0$ on $(-7, -2)$ so $f(x)$ is decreasing on this interval.