Question

calculus question: find a function f such that the graph of f has a horizontal tangent at (2,0) and f''(x)=2x.

Answer 1

will you give medal

Answer 2

yes

Answer 3

ok

Answer 4

can you help me please?

Answer 5

yeah but do you have a simple pic i can see of it it don't make scene but i can still help

Answer 6

@Studentc14

Answer 7

There's no picture it's just words on my homework

Answer 8

hmm

f''=2x
so wouldmt f'(x)= x^2

Answer 9

ah i think i got it

Answer 10

\[f"(x) =2x\]
\[f'(x)=x^2+consant\]
\[f'(2)=2^2+constant=0\]

constant would then = -4

Answer 11

hmm

Answer 12

i think the original function is
\[f(x)=\frac{ x^3 }{ 3 }-4\]

Answer 13

-4x i mean

Answer 14

ya thats it

Answer 15

see what i did there?

Answer 16

\[d(\frac{ x^3 }{ 3 } - 4x = x^2-4\]

then the derivative of that is 2x

Answer 17

ah yes i see! thank you @TylerD

Answer 18

this is a derivative + integrals problem. and im not to good with integrals but was able to solve it by eyeballing it basically without taking the integral

Answer 19

yeah i see, thank you! would it also have +c at the end after -4x or no? @TylerD

Answer 20

no constant was just the placeholder so i know theres something there then solved for it

Answer 21

actually it says (2,0) so wait a min here

Answer 22

i mean in the final solution, because couldn't x^3/3 -4x still be shifted up or down and have the same derivatives?

Answer 23

meh i tink thats right though

Answer 24

ok thank you

Answer 25

ya it could be shifted

Answer 26

it would be shifted up by like +5.333

Answer 27

ok thanks!

Answer 28

\[f''(x)=2x \\ f'(x)=x^2+b \\ f(x)=\frac{x^3}{3}+bx+c\]
so we know f looks like that above
we want to find the tangent to f such that there is a horizontal tangent at (2,0)
This also means (2,0) belongs to the function f
so f'(x)=0 when x=2
so b=4 is right
\[f(x)=\frac{x^3}{3}+4x+c \\ \text{ then use (2,0) \to find c } f(2)=\frac{8}{3}+8+c \\0=\frac{8}{3}+8+c \text{ since } f(2)=0 \\ 0=\frac{8}{3}+\frac{24}{3}+c\]
then solve for c

Answer 29

oops b=-4

Answer 30

\[f(x)=\frac{x^3}{3}-4x+c \\ f(2)=\frac{8}{3}-8+c \\ 0=\frac{8}{3}-\frac{24}{3}+c \\\]

Answer 31

thank you!! @freckles