Question

hi I don’t know how to answer part B of the Question, I’m in high school calculus one, and this is a homework

Answer 1

The given function is

`f(x)=7x-3x^2`

Using the limit definition

`f^(\prime)(x)=\lim _(h\to0)(f(x+h)-f(x))/(h)`

Substitute x+h for x

This gives

`f^(\prime)(x)=\lim _(h\to0)(7(x+h)+3(x+h)^2-(7x-3x^2))/(h)`

Simplify the limit

`\begin{gathered} f^(\prime)(x)=\lim _(h\to0)\frac{7x+7h-3(x^2+2hx+h^2)^{}-7x+3x^2}{h} \\ \end{gathered}`

This further gives

`f^(\prime)(x)=\lim _(h\to0)(7x+7h-3x^2-6hx-3h^2-7x+3x^2)/(h)`

Simplify further

`f^(\prime)(x)=\lim _(h\to0)(7h-6hx-3h^2)/(h)`

Simplify the fraction

`\begin{gathered} f^(\prime)(x)=\lim _(h\to0)((7h)/(h)-(6hx)/(h)-(3h^2)/(h)) \\ f^(\prime)(x)=\lim _(h\to0)(7-6x-3h) \end{gathered}`

Find the limit

`\begin{gathered} f^(\prime)(x)=(7-6x-3(0)) \\ f^(\prime)(x)=7-6x \end{gathered}`

Therefore, the solution is

`f^(\prime)(x)=7-6x`