If fx- 3x-1- 0- x - 2 1-9x- 2- x- 3 30-2x- x- 3 find f52

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Standard XII

Mathematics

Derivative from First Principle

If fx= 3x+1...

Question

If $f (x) = ⎧ ⎨ ⎩ \begin{matrix} 3 x + 1, 0 \leq x \leq 2 1 + 9 x, 2 < x < 3 30 + 2 x, x \geq 3 \end{matrix}$ find
$f (\frac{5}{2})$

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Solution

Given,
$f (\frac{5}{2}) = f (2.5)$
We have to find $f (2.5)$ so we use function $1 + 9 x$ because it define from $2 < x < 3 a n d 2.5$ lie in this interval.
Therefore , put $x = 2.5$ in $f n 1 + 9 x$
$f (2.5) = 1 + 9 (2.5)$
$f (2.5) = 1 + 22.5$
$f (2.5) = 23.5$

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If f(x)=⎧⎨⎩3x+1, 0≤x≤21+9x, 2<x<330+2x, x≥3 find f(52)

Given, f(52)=f(2.5)We have to find f(2.5) so we use function 1+9x because it define from 2<x<3 and 2.5 lie in this interval.Therefore , put x=2.5 in fn1+9xf(2.5)=1+9(2.5)f(2.5)=1+22.5f(2.5)=23.5

If $f (x) = ⎧ ⎨ ⎩ \begin{matrix} 3 x + 1, 0 \leq x \leq 2 1 + 9 x, 2 < x < 3 30 + 2 x, x \geq 3 \end{matrix}$ find
$f (\frac{5}{2})$

Given,
$f (\frac{5}{2}) = f (2.5)$
We have to find $f (2.5)$ so we use function $1 + 9 x$ because it define from $2 < x < 3 a n d 2.5$ lie in this interval.
Therefore , put $x = 2.5$ in $f n 1 + 9 x$
$f (2.5) = 1 + 9 (2.5)$
$f (2.5) = 1 + 22.5$
$f (2.5) = 23.5$