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What is the number of points of local minima of the function $\mathrm{f}(\mathrm{x}) ?$
Consider the function $\mathrm{f}(\mathrm{x})=(\mathrm{x}-1)^{2}(\mathrm{x}+1)(\mathrm{x}-2)^{3}$
Options:
Consider the function $\mathrm{f}(\mathrm{x})=(\mathrm{x}-1)^{2}(\mathrm{x}+1)(\mathrm{x}-2)^{3}$
Solution:
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Verified Answer
The correct answer is:
Two
$f(x)=(x-1)^{2}(x+1)(x-2)^{3}$
$f^{\prime}(x)=2(x-1)(x+1)(x-2)^{3}+(x-1)^{2}(x-2)^{3}$
$+(x-1)^{2}(x+1) 3(x-2)^{2}$
$$
\begin{aligned}(x-1)(x-2)^{2}[2(x+1)(x-2)+(x-1)(x-2)\\+3(x-1)(x+1)] \end{aligned}
$$
$f^{\prime}(x)=(x-1)(x-2)^{2}\left[2 x^{2}-2 x-4+x^{2}-3 x+2\right.$
$\left.+3 x^{2}-3\right]$
$=(x-1)(x-2)^{2}\left[6 x^{2}-5 x-5\right]$
For maxima and minima $f^{\prime}(x)=0$
$(x-1)(x-2)^{2}\left[6 x^{2}-5 x-5\right]=0$
$x=1,2,2, \frac{5 \pm \sqrt{145}}{12}$ The change in signs of $\mathrm{f}^{\prime}(\mathrm{x})$ for
diffrent values of $x$ is shown:
$\therefore$ Local Minima are
$x=\frac{5-\sqrt{145}}{12} \& x=\frac{5+\sqrt{145}}{12}$
$f^{\prime}(x)=2(x-1)(x+1)(x-2)^{3}+(x-1)^{2}(x-2)^{3}$
$+(x-1)^{2}(x+1) 3(x-2)^{2}$
$$
\begin{aligned}(x-1)(x-2)^{2}[2(x+1)(x-2)+(x-1)(x-2)\\+3(x-1)(x+1)] \end{aligned}
$$
$f^{\prime}(x)=(x-1)(x-2)^{2}\left[2 x^{2}-2 x-4+x^{2}-3 x+2\right.$
$\left.+3 x^{2}-3\right]$
$=(x-1)(x-2)^{2}\left[6 x^{2}-5 x-5\right]$
For maxima and minima $f^{\prime}(x)=0$
$(x-1)(x-2)^{2}\left[6 x^{2}-5 x-5\right]=0$
$x=1,2,2, \frac{5 \pm \sqrt{145}}{12}$ The change in signs of $\mathrm{f}^{\prime}(\mathrm{x})$ for
diffrent values of $x$ is shown:
$\therefore$ Local Minima are
$x=\frac{5-\sqrt{145}}{12} \& x=\frac{5+\sqrt{145}}{12}$
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