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find all intervals on which $f(x)$ is decreasing. $f(x)=x^{4}-12x^{3}-1…

Question

find all intervals on which $f(x)$ is decreasing.
$f(x)=x^{4}-12x^{3}-19$

Explanation:

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

Using the power rule \((x^n)^\prime=nx^{n - 1}\), for \(f(x)=x^{4}-12x^{3}-19\), we have \(f^\prime(x)=4x^{3}-36x^{2}=4x^{2}(x - 9)\)

Step2: Set \(f^\prime(x)<0\) to find where \(f(x)\) is decreasing

Since \(4x^{2}\geq0\) for all real \(x\) (equality holds when \(x = 0\)), the sign of \(f^\prime(x)\) is determined by \((x - 9)\).
We want \(f^\prime(x)=4x^{2}(x - 9)<0\).
Because \(4x^{2}>0\) for \(x
eq0\), we solve \(x-9<0\) (excluding \(x = 0\) does not affect the inequality since \(f^\prime(0)=0\)).

Answer:

The function \(f(x)\) is decreasing on the interval \((-\infty,9)\)