+26367 Simplify f(x+h)-f(x)/ h
where (f)x=2x2 + x
\(\begin{array}{rcll} f(x) &=& 2x^2+x\\ f(x+h)&=&2(x+h)^2+(x+h)\\\\ \text{The difference quotient is:}\\ \dfrac{\Delta y}{\Delta x} &=& \dfrac{f(x+h)-f(x)}{h}\\ &=& \dfrac{2(x+h)^2+(x+h)-(2x^2+x)}{h}\\ &=& \dfrac{2(x+h)^2+ x +h-2x^2 - x }{h}\\ &=& \dfrac{2(x+h)^2 +h-2x^2 }{h}\\ &=& \dfrac{2(x^2+2xh+h^2) +h-2x^2 }{h}\\ &=& \dfrac{2x^2+4xh+2h^2 +h-2x^2 }{h}\\ &=& \dfrac{ 4xh+2h^2 +h }{h}\\ &=& \dfrac{ h( 4x+2h +1) }{h}\\ \dfrac{\Delta y}{\Delta x} &=& 4x+2h +1 \end{array} \)
\(\begin{array}{rcll} f'(x) &=& \dfrac{dy}{dx} \\ \dfrac{dy}{dx} &=& \lim \limits_{h\to 0} { \left( 4x+2h +1 \right) } \\ \dfrac{dy}{dx} &=& \left( 4x+2\cdot 0 +1 \right) \\ \dfrac{dy}{dx} &=& \left( 4x+1 \right) \\\\ \mathbf{f'(x)} &\mathbf{=}& \mathbf{4x+1} \end{array}\)
