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To determine where the function is strictly decreasing, we first need to find its derivative, $f'(x)$. $$f(x) = x^4 - 4x^3 + 10$$ $$f'(x) = 4x^3 - 12x^2$$
To find the critical points, we set the derivative equal to zero and solve for $x$. $$4x^3 - 12x^2 = 0$$ $$4x^2(x - 3) = 0$$ This gives us $x = 0$ and $x = 3$ as critical points.
Now, we need to test the intervals determined by the critical points to see where $f'(x)$ is negative (strictly decreasing). The intervals are $(-\infty, 0)$, $(0, 3)$, and $(3, \infty)$. Choose test points in each interval: - For $(-\infty, 0)$, let $x = -1$. Then $f'(-1) = 4(-1)^3 - 12(-1)^2 = -4 - 12 = -16 < 0$. So, $f(x)$ is strictly decreasing on $(-\infty, 0)$. - For $(0, 3)$, let $x = 1$. Then $f'(1) = 4(1)^3 - 12(1)^2 = 4 - 12 = -8 < 0$. So, $f(x)$ is strictly decreasing on $(0, 3)$. - For $(3, \infty)$, let $x = 4$. Then $f'(4) = 4(4)^3 - 12(4)^2 = 4(64) - 12(16) = 256 - 192 = 64 > 0$. So, $f(x)$ is strictly increasing on $(3, \infty)$.
Since $f'(x) < 0$ on $(-\infty, 0)$ and $(0, 3)$, the function is strictly decreasing on these intervals. Combining these intervals, we can say that the function is strictly decreasing on $(-\infty, 3)$, except at $x=0$ where $f'(0)=0$. However, since the question asks for the interval where the function is strictly decreasing, we can consider the interval $(-\infty, 3]$. But since the function is not strictly decreasing at $x=3$, the interval is $(-\infty, 3)$.
Final Answer: $(-\infty, 3)$
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