Applications of Derivative — MATHS STD 12 Science — Question
Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSApplications of Derivative3 Marks
Question
In interval $[ 1 , 5]$ Find the absolute maximum and absolute minimum values of given function $f(x)=x^2-4 x+8$.
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Answer
We know :$ \begin{aligned} f(x) & =x^2-4 x+8 \\ \text { or } \quad f^{\prime}(x) & =2 x-4=2(x-2) \end{aligned} $ Put $f^{\prime}(x)=0$ we get $x=2$ We find the values of $f$ at $x=1, x=2$, and $x=5$. now$ \begin{aligned} \therefore f(x) & =x^2-4 x+8 \\ \therefore f(1) & =(1)^2-4 \times 1+8=5 \\ f(2) & =(2)^2-4 \times 2+8=4 \\ f(5) & =(5)^2-4 \times 5+8=13 \end{aligned} $ Thus we find that in $[1,5]$ function at $x=5$ absolute maximum value is 13 and at $x=2$ absolute minimum value is 4 .
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