If $f (x) = m x + c$ and if $f (0) = f^{1} (0) = 1$ then find $f (2)$

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Solution

$f (x) = m x + c f^{'} (x) = \frac{d f (x)}{d x} = \frac{d (m x + c)}{d x} = m f^{'} (0) = 1 \Rightarrow m = 1 f (0) = 1 \Rightarrow (1 \times 0) + c = 1 \Rightarrow c = 1 f (x) = x + 1 \Rightarrow f (2) = 1 \times 2 + 1 = 3$

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If f(x)=mx+c and if f(0)=f1(0)=1 then find f(2)

f(x)=mx+cf′(x)=df(x)dx=d(mx+c)dx=mf′(0)=1⇒m=1f(0)=1⇒(1×0)+c=1⇒c=1f(x)=x+1⇒f(2)=1×2+1=3

If $f (x) = m x + c$ and if $f (0) = f^{1} (0) = 1$ then find $f (2)$

$f (x) = m x + c f^{'} (x) = \frac{d f (x)}{d x} = \frac{d (m x + c)}{d x} = m f^{'} (0) = 1 \Rightarrow m = 1 f (0) = 1 \Rightarrow (1 \times 0) + c = 1 \Rightarrow c = 1 f (x) = x + 1 \Rightarrow f (2) = 1 \times 2 + 1 = 3$