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Increasing and Decreasing Functions — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIncreasing and Decreasing Functions5 Marks
Question
Determine the values of $x$ for which the function $f(x) = x^2 - 6x + 9$ is increasing or decreasing. Also, find the coordinates of the point on the curve $y = x^2 - 6x + 9$ where the normal is parallel to the line $y = x + 5.$
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Answer

We have, $f(x) = x^2 - 6x + 9$
$\therefore f'(x) = 2x - 6$
Critical points $f'(x) = 0 $
$\Rightarrow 2(x - 3) = 0 $
$\Rightarrow x = 3$
Clearly, $f'(x) > 0$
if $x > 3$
$f'(x) < 0$
if $x < 3$
Thus, $f(x)$ is increases in $(3,\infty),$ decreases in $(-\infty,3)$
II part:
The given equation of curves
$y = x^2 - 6x + 9 ....(i) $
$y = x + 5 ....(ii)$
Slope of $(i)$
$\text{m}_1=\frac{\text{dy}}{\text{dx}}=2\text{x}-6$
Slope of $(ii)$
$\text{m}_2=1$
Given that slope of normal to $(i)$ is parallel to $(ii)$
$\therefore\ \frac{-1}{2\text{x}-6}=1$
$\Rightarrow2\text{x}-6=-1$
$\Rightarrow\text{x}=\frac{5}{2}$
From $(i)$
$\text{y}=\frac{25}{4}-15+9$
$=\frac{25}{4}-6$
$=\frac{1}{4}$
​​​​​​​Thus, the required point is $\Big(\frac{5}{2},\frac{1}{4}\Big).$

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