The function $f(x) = \frac{-3}{4}x^4 - 8x^3 - \frac{45}{2}x^2 + 163$ has a local maxima at

Given function: $f(x) = \frac{-3}{4}x^4 - 8x^3 - \frac{45}{2}x^2 + 163$

The first derivative is:

$f'(x) = \frac{-3}{4} \times 4x^3 - 8 \times 3x^2 - \frac{45}{2} \times 2x$

$f'(x) = -3x^3 - 24x^2 - 45x$

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To find critical points, set $f'(x) = 0$:

$-3x^3 - 24x^2 - 45x = 0$

$-3x(x^2 + 8x + 15) = 0$

This gives either $x = 0$ or $x^2 + 8x + 15 = 0$

Factoring the quadratic:

$(x + 3)(x + 5) = 0$

$x = -3$ or $x = -5$

Critical points: $x = 0, -3, -5$

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The second derivative is:

$f''(x) = -9x^2 - 48x - 45$

Testing each critical point:

At $x = 0$:

$f''(0) = -9(0)^2 - 48(0) - 45$

$f''(0) = -45 < 0$

Therefore $x = 0$ is a local maximum.

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At $x = -3$:

$f''(-3) = -9(9) - 48(-3) - 45$

$f''(-3) = -81 + 144 - 45$

$f''(-3) = 18 > 0$

Therefore $x = -3$ is a local minimum.

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At $x = -5$:

$f''(-5) = -9(25) - 48(-5) - 45$

$f''(-5) = -225 + 240 - 45$

$f''(-5) = -30 < 0$

Therefore $x = -5$ is a local maximum.

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The function has local maxima at $x = 0$ and $x = -5$.