If f (x) = x^2+ 1 then f (x+h) - f (x) over h

asked by guest
on Jan 24, 2025 at 1:49 am



You asked:

Evaluate the function \(f\left( x \right) = {x}^{2} + 1\), \(f\left( x + h \right) - \frac{f\left( x \right)}{h}\).

MathBot Answer:

$$\begin{aligned}f\left( x + h \right) - \frac{f\left( x \right)}{h}&=\frac{h^{3} + 2 h^{2} x + x^{2} h + h - x^{2} - 1}{h}\end{aligned}$$

$$\begin{aligned}f\left( x + h \right) - \frac{f\left( x \right)}{h}&=1 + \left(h + x\right)^{2} - \frac{1 + x^{2}}{h}\\f\left( x + h \right) - \frac{f\left( x \right)}{h}&=\frac{- x^{2} - 1}{h} + 1 + \left(h + x\right)^{2}\\f\left( x + h \right) - \frac{f\left( x \right)}{h}&=\frac{- x^{2} - 1}{h} + 1 + h^{2} + 2 h x + x^{2}\\f\left( x + h \right) - \frac{f\left( x \right)}{h}&=\frac{- x^{2} - 1}{h} + h^{2} + 2 h x + x^{2} + 1\\f\left( x + h \right) - \frac{f\left( x \right)}{h}&=\frac{h^{3} + 2 h^{2} x + h x^{2} + h - x^{2} - 1}{h}\\f\left( x + h \right) - \frac{f\left( x \right)}{h}&=\frac{h^{3} + 2 h^{2} x + x^{2} h + h - x^{2} - 1}{h}\\\end{aligned}$$

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