Question

What is the value of f ′(5) for f(x) = 3x^2 – 2x – 4?

I don't want the answer, but just help with the steps of setting up the problem please

Answer 1

first take the derivative of the function and then substitute 5 for x

Answer 2

\[\huge\rm f(x) = \color{blue}{3x^2 – 2x – 4}\]
what would be the derivative of this function ??

Answer 3

what equation do you use to find the derivative? f(x+h)-f(x)/h ?

Answer 4

@Nnesha

Answer 5

same equation
for f(x+h) replace `x` with `x+h`

Answer 6

oh no the given function f(x) = 3x^2 – 2x – 4

Answer 7

so 3(5+h)^2-2(5+h)-4 -(3(5)^2-2(5)-4)/h ?

Answer 8

could I also use this formula? f(x)-f(a)/x-a ? or is that for something entirely different?

Answer 9

we have to find the value of f'(5) where x=5 so
\[\huge\rm \frac{ f(\color{Red}{x}+h)-f(\color{Red}{x}) }{ h }\]
\[\huge\rm \frac{ f(\color{Red}{5}+h)-f(\color{Red}{x}) }{ h }\]
to find the value of f'(5) we have to take the limit as x approaches 0
\[\large\rm f'(5)=\lim_{h \rightarrow 0}\frac{ f(\color{Red}{5}+h)-f(\color{Red}{x}) }{ h }\]

Answer 10

correction ***
\[\huge\rm \frac{ f(\color{Red}{5}+h)-f(\color{Red}{5}) }{ h }\]
and i meant to say `h approaches 0`

Answer 11

ooooh okay, so find out what 3(5)^2-2(5)-4 =, 36, and then it looks like (36+h)-36 / h?

Answer 12

to find the derivative of a function at some point (f'(a))
first step) find `f(a)`
2nd step) find `f(a+h)`
3rd) use the formula \[\frac{ f(a+h)-f(a) }{ h }\] and then take the limit as h->0

Answer 13

hmm i didn't get 36..

Answer 14

oops, 61? must have multiplied wrong

Answer 15

yeah looks good
now find f(a+h) for your question a=5
so substitute `5+h` into the original equation

Answer 16

\[\large\rm f'(5)= \lim_{h \rightarrow 0} \frac{ f(\color{blue}{5+h})- f(\color{Red}{a}) }{ h }\]\[\large\rm f'(5)= \lim_{h \rightarrow 0} \frac{[\color{blue}{3(5+h)^2-2(5+h)-4} ]- \color{Red}{61} }{ h }\] simplify :=))

Answer 17

confused ??? ~o.0~

Answer 18

tag me if u have question abt it :=))

Answer 19

I believe the answer is 28 @Nnesha very confused though haha but you've helped a lot!

Answer 20

confused about what ?

Answer 21

\[\large\rm \lim_{h\rightarrow 0 }\frac{ 3(25+10h+h^2)-10-2h -61}{ h}\]\[\large\rm \lim_{h\rightarrow 0 }\frac{ 75+30h+3h^2-10-2h -61}{ h}\]
simplify
the h canceled out (at the denominator )
then substitute h for 0 you will find the value of f'(5)
:=)) good job!

Answer 22

The approach you're using here is called "definition of the derivative." Were you told to use this method?

If you have already learned some of the basic derivative rules, try tackling this problem with them. Pls. show your work.