## Calculus with Applications (10th Edition)

$$f(x)=-4 x^{2}+11 x$$ Now use the rules for limits to get $$\begin{split} f^{\prime}(x) &=\lim\limits_{h \to 0} \frac{f(x +h)- f(x)}{h}\\ &=\lim\limits_{h \to 0}\frac{-4 (x +h)^{2}+11 (x +h)-(-4 x^{2}+11 x) }{h}\\ &=\lim\limits_{h \to 0}\frac{-8 x h-4 h^{2}+11 h}{h}\\ &=\lim\limits_{h \to 0}(-8 x -4 h+11)\\ &=-8 x +11. \end{split}$$ Use this result to find $$f^{\prime}(2)=-8 (2) +11=-5,$$ $$f^{\prime}(16)=-8 (16) +11=-117$$ and $$f^{\prime}(-3)=-8 (-3) +11=35$$
$$f(x)=-4 x^{2}+11 x$$ Go through the four steps to find $f^{\prime}(x)$ Step 1 Find $f(x+h)$ by replacing $x$ with $x +h$ $$f(x +h)=-4 (x +h)^{2}+11 (x +h)$$ Step 2 $$f(x +h) f(x)=-4 (x +h)^{2}+11 (x +h)-(-4 x^{2}+11 x)$$ Step 3 $$\frac{f(x +h)- f(x)}{h}= \frac{-4 (x +h)^{2}+11 (x +h)-(-4 x^{2}+11 x) }{h}$$ Step 4 Now use the rules for limits to get $$\begin{split} f^{\prime}(x) &=\lim\limits_{h \to 0} \frac{f(x +h)- f(x)}{h}\\ &=\lim\limits_{h \to 0}\frac{-4 (x +h)^{2}+11 (x +h)-(-4 x^{2}+11 x) }{h}\\ &=\lim\limits_{h \to 0}\frac{-8 x h-4 h^{2}+11 h}{h}\\ &=\lim\limits_{h \to 0}(-8 x -4 h+11)\\ &=-8 x +11. \end{split}$$ Use this result to find $$f^{\prime}(2)=-8 (2) +11=-5,$$ $$f^{\prime}(16)=-8 (16) +11=-117$$ and $$f^{\prime}(-3)=-8 (-3) +11=35.$$