Question

ONE QUESTION .. WILL MEDAL!

Answer 1

I really need help on this question ^

Answer 2

I think they meant to "use" f(x) in the rational expression

thus

Answer 3

well i cant get anyone who actually knows how to do this question so I hope you do dude
its a hard one

Answer 4

no. its derivative

Answer 5

\(\bf \cfrac{f({\color{brown}{ a}}+h)-f({\color{brown}{ a}})}{h}\qquad f({\color{brown}{ x}})={\color{brown}{ x}}^2+5{\color{brown}{ x}}\quad or\quad f({\color{brown}{ a}})={\color{brown}{ a}}^2+5{\color{brown}{ a}}
\\ \quad \\

f({\color{blue}{ a+h}})=({\color{blue}{ a+h}})^2+5({\color{blue}{ a+h}})\qquad thus
\\ \quad \\
\cfrac{f({\color{brown}{ a}}+h)-f({\color{brown}{ a}})}{h}\implies \cfrac{({\color{blue}{ a+h}})^2+5({\color{blue}{ a+h}})\quad -\quad (a^2+5a)}{h}\)

Answer 6

recal that \[limx rightarrowh={ h(a+h)+f(a)}/h\]

Answer 7

hmm ok

Answer 8

very correct @jdoe0001

Answer 9

that looks like C

Answer 10

expand the squared expression, and cancel out what you can, see what you end up with

Answer 11

ok one sec

Answer 12

so what exactly am i solving.. sorry

Answer 13

\(\bf \cfrac{f({\color{brown}{ a}}+h)-f({\color{brown}{ a}})}{h}\qquad f({\color{brown}{ x}})={\color{brown}{ x}}^2+5{\color{brown}{ x}}\quad or\quad f({\color{brown}{ a}})={\color{brown}{ a}}^2+5{\color{brown}{ a}}
\\ \quad \\

f({\color{blue}{ a+h}})=({\color{blue}{ a+h}})^2+5({\color{blue}{ a+h}})\qquad thus
\\ \quad \\
\cfrac{f({\color{brown}{ a}}+h)-f({\color{brown}{ a}})}{h}\implies \cfrac{({\color{blue}{ a+h}})^2+5({\color{blue}{ a+h}})\quad -\quad (a^2+5a)}{h}\)

I think the confusing part is that, they used "x" on the f(x) function
and they use f(a) in the part where it is going to be used

but in short, you can just replace the "x" for "a" and use \(\bf f({\color{brown}{ a}})={\color{brown}{ a}}^2+5{\color{brown}{ a}}\)

Answer 14

dang.. ok one sec

Answer 15

ok um one more sec

Answer 16

what does the binomial \(\large \bf (a+h)^2\) give you?

Answer 17

almost done one sec

Answer 18

k