# Differentiating to find one unknown function out of three?

mimi.janson
differentiating to find one unknown function out of three??

## Homework Statement

Hi everyone

I need some help with a question that i have solved yet i find it hard to understand.

I have three given functions in the picture attached. All three consist of one single graph. And they give me the functions for two parts whereas i have to find out the last one which is the parable.

## Homework Equations

I needed to differentiate what i already had given, since i wanted to find the slope of the functions since i also know that where one function stops and the other begins i have the same slope.

## The Attempt at a Solution

I tried to solve it by saying that g(x)=ax^2+bx+c

then i differentiate it and it becomes:

g'(x)=2ax+b

i now look at the first function f(x)=-1 is f^' (x)=0 this means that the slope for this function is 0.

the slope for the points (-2;0) must therefore also be 0.

i put this in the parable function

g'(x)=2ax+b
g'(-2)=-4a+b=0

then i know that in the points (0;0) the slope of the parable must be the same as the one for h(x)

i have to find h'(0) and it won't work unless i get it to be h'(0)=1=b but i don't get 1 since i say h'(0)=0,006*0^2-0,18
h'(0)=-0,18

the end result should be g(x) = 1/4 x + x

what is wrong where i stop up ? can someone please tell my why i won't get it right..

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Homework Helper

This is very hard to follow. Could you please post the question you're trying to answer in its entirety? And it's called a parabola, not a parable.

Homework Helper
Dearly Missed

## Homework Statement

Hi everyone

I need some help with a question that i have solved yet i find it hard to understand.

I have three given functions in the picture attached. All three consist of one single graph. And they give me the functions for two parts whereas i have to find out the last one which is the parable.

## Homework Equations

I needed to differentiate what i already had given, since i wanted to find the slope of the functions since i also know that where one function stops and the other begins i have the same slope.

## The Attempt at a Solution

I tried to solve it by saying that g(x)=ax^2+bx+c

then i differentiate it and it becomes:

g'(x)=2ax+b

i now look at the first function f(x)=-1 is f^' (x)=0 this means that the slope for this function is 0.

the slope for the points (-2;0) must therefore also be 0.

i put this in the parable function

g'(x)=2ax+b
g'(-2)=-4a+b=0

then i know that in the points (0;0) the slope of the parable must be the same as the one for h(x)

i have to find h'(0) and it won't work unless i get it to be h'(0)=1=b but i don't get 1 since i say h'(0)=0,006*0^2-0,18
h'(0)=-0,18

the end result should be g(x) = 1/4 x + x

what is wrong where i stop up ? can someone please tell my why i won't get it right..

I do not see any attached picture.

I am assuming that when you say "i", you are referring to yourself rather than the square root of -1. I guess you mean "I".

I find it impossible to follow what you are doing.

RGV

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