hw 7 - definition of the derivative section 2.2: problem 5 (1 point)\nlet $f(x)=5x + 3x^{2}$. if $h\neq0$…

hw 7 - definition of the derivative section 2.2: problem 5 (1 point)\nlet $f(x)=5x + 3x^{2}$. if $h\neq0$, then the difference quotient can be simplified as\n\\(\\frac{f(x + h)-f(x)}{h}=ah + bx + c\\),\nwhere $a$, $b$, and $c$ are constants. (note: its possible for one or more of these constants to be 0.) find the constants.\n$a = \\square$, $b = \\square$, and $c = \\square$.\nuse the simplified expression to find $f(x)=\\lim_{h\\to0}\\frac{f(x + h)-f(x)}{h}=\\square$.\nfinally, find each of the following.\n$f(1)=\\square$, $f(2)=\\square$, and $f(3)=\\square$.

hw 7 - definition of the derivative section 2.2: problem 5 (1 point)\nlet $f(x)=5x + 3x^{2}$. if $h\neq0$, then the difference quotient can be simplified as\n\\(\\frac{f(x + h)-f(x)}{h}=ah + bx + c\\),\nwhere $a$, $b$, and $c$ are constants. (note: its possible for one or more of these constants to be 0.) find the constants.\n$a = \\square$, $b = \\square$, and $c = \\square$.\nuse the simplified expression to find $f(x)=\\lim_{h\\to0}\\frac{f(x + h)-f(x)}{h}=\\square$.\nfinally, find each of the following.\n$f(1)=\\square$, $f(2)=\\square$, and $f(3)=\\square$.

Answer

Explanation:

Step1: Find (f(x + h))

[ \begin{align*} f(x+h)&=5(x + h)+3(x + h)^2\ &=5x+5h + 3(x^{2}+2xh+h^{2})\ &=5x+5h+3x^{2}+6xh + 3h^{2} \end{align*} ]

Step2: Calculate (f(x + h)-f(x))

[ \begin{align*} f(x + h)-f(x)&=(5x+5h+3x^{2}+6xh + 3h^{2})-(5x + 3x^{2})\ &=5h+6xh+3h^{2} \end{align*} ]

Step3: Find the difference - quotient (\frac{f(x + h)-f(x)}{h})

[ \begin{align*} \frac{f(x + h)-f(x)}{h}&=\frac{5h+6xh+3h^{2}}{h}\ &=3h + 6x+5 \end{align*} ] Comparing with (Ah + Bx + C), we have (A = 3), (B = 6), (C = 5)

Step4: Find (f^{\prime}(x))

[ \begin{align*} f^{\prime}(x)&=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}\ &=\lim_{h\rightarrow0}(3h + 6x+5)\ &=6x + 5 \end{align*} ]

Step5: Calculate (f^{\prime}(1)), (f^{\prime}(2)) and (f^{\prime}(3))

[ \begin{align*} f^{\prime}(1)&=6\times1+5=11\ f^{\prime}(2)&=6\times2+5=17\ f^{\prime}(3)&=6\times3+5=23 \end{align*} ]

Answer:

(A = 3), (B = 6), (C = 5), (f^{\prime}(x)=6x + 5), (f^{\prime}(1)=11), (f^{\prime}(2)=17), (f^{\prime}(3)=23)