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Question Number 194085 by 073 last updated on 27/Jun/23

$$\mathrm{f}\left(\mathrm{x}\right)-\mathrm{f}(\mathrm{x}-1)=\mathrm{x}+1$$$${\mathrm{f}}^{\prime }\left(6\right)=?$$$$\mathrm{solution}??$$

Answered by qaz last updated on 27/Jun/23

$$f\left(x\right)={ax}^2+bx+c$$$$f(x-1)=a{(x-1)}^2+b(x-1)+c$$$$f\left(x\right)-f(x-1)=a(2x-1)+b=x+1$$$$\Rightarrow a=\frac{1}{2}b=\frac{3}{2}$$$$\Rightarrow f\left(x\right)=\frac{1}{2}{x}^2+\frac{3}{2}x+c\Rightarrow f{\left(6\right)}^{\prime }=\frac{15}{2}$$

Answered by mr W last updated on 27/Jun/23

$$a_k-a_{k-1}=k+1$$$$\underset{k=1}{\sum ^n}(a_k-a_{k-1})=\underset{k=1}{\sum ^n}(k+1)$$$$a_n-a_0=\frac{n(n+3)}{2}$$$$a_n=\frac{n(n+3)}{2}+a_0$$$$\Rightarrow f\left(x\right)=\frac{x(x+3)}{2}+f\left(0\right)$$$$\Rightarrow f^{\prime }\left(x\right)=\frac{2x+3}{2}$$$$\Rightarrow f^{\prime }\left(6\right)=\frac{15}{2}$$

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