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To find the critical points, we first need to find the first derivative of the function $f(x)$. $$f'(x) = \frac{d}{dx}(4x^{3}-18x^{2}+27x-7) = 12x^{2} - 36x + 27$$
To find the critical points, we set the first derivative equal to zero and solve for $x$. $$12x^{2} - 36x + 27 = 0$$ Divide by 3: $$4x^{2} - 12x + 9 = 0$$ $$(2x - 3)^{2} = 0$$ $$2x - 3 = 0$$ $$x = \frac{3}{2}$$ So, there is only one critical point at $x = \frac{3}{2}$.
To determine whether the critical point is a maximum, minimum, or neither, we find the second derivative of the function. $$f''(x) = \frac{d}{dx}(12x^{2} - 36x + 27) = 24x - 36$$
Now, we evaluate the second derivative at the critical point $x = \frac{3}{2}$. $$f''\left(\frac{3}{2}\right) = 24\left(\frac{3}{2}\right) - 36 = 36 - 36 = 0$$ Since the second derivative is zero at the critical point, the second derivative test is inconclusive.
We need to analyze the sign of the first derivative around $x = \frac{3}{2}$. Since $f'(x) = (2x - 3)^{2}$, $f'(x)$ is always non-negative. For $x < \frac{3}{2}$, $f'(x) > 0$. For $x > \frac{3}{2}$, $f'(x) > 0$. Since the sign of the first derivative does not change around $x = \frac{3}{2}$, there is neither a maximum nor a minimum at this point. The function is increasing on both sides of $x = \frac{3}{2}$.
Therefore, the function $f(x) = 4x^{3} - 18x^{2} + 27x - 7$ has neither maxima nor minima.
Final Answer: The function has neither maxima nor minima.
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