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Q.
Find the points of local minima for the function $f\left(x\right) = -\frac{3}{4}x^{4} - 8x^{3} - \frac{45}{2}x^{2} + 105$
Application of Derivatives
Solution:
We have,
$f\left(x\right) = -\frac{3}{4}x^{4} - 8x^{3} - \frac{45}{2}x^{2} + 105$
$\Rightarrow f'\left(x\right) = - 3x^{3} - 24x^{2} - 45x$
$ = - 3x\left(x^{2} + 8x + 15\right)$
For local maximum or local minimum, we must have $f'\left(x\right) = 0$
$\Rightarrow - 3x\left(x^{2} + 8x + 15\right) = 0$
$\Rightarrow 3x\left(x + 3\right) \left(x + 5\right) = 0$
$\Rightarrow x = 0$, $- 3$, $-5$
Thus, $x = 0$, $x = - 3$ and $x = - 5$ are the possible points of local maxima or minima.
Now we test the function at each of these points.
We have, $f''(x) = - 9x^2 - 48 x - 45$
At $x = 0$ : We have,
$f''(0) = - 45 < 0$
So, $x = 0$ is point of local maximum.
At $x = - 3$ : We have,
$f''(- 3) = - 9(-3)^2 - 48(-3) - 45 = 18 > 0$
So, $x = - 3$ is a point of local minimum.
At $x = - 5$ : We have,
$f''( - 5) = - 9(-5)^2 - 48(-5) - 45 = -30 < 0$
So, $x = - 5$ is point of local maximum.