1. ## [SOLVED] Specific Gradient of a curve

Sketch the graph of y=(x+2)(X+1)(X-1) = X^3+2x^2-x-2 in the range -2.5< or equal to X < or equal to 1.5. (You must determine the points where the gradient is zero and where the curve crosses the x and y Axes.

I have substituted the values of x between the stated ranges to get my graph, I can then clearly see where the graph cross the x and Y axis.

I then have to find where the gradiet is zero, This is where my problem lies, I can see on the graph roughly where it will be, however i need the exact answer. All the books and example i have been able to find just show how to find the gradient at a specific point not a specific gradient.

I know that i haveto differentiate to get the gradient, but after that i am a little bit stuck.

2. Hello,

I have substituted the values of x between the stated ranges to get my graph, I can then clearly see where the graph cross the x and Y axis.
I think you have to solve them algebraically ^^'
That is f(0)=... ? and f(x)=0 --> x=... ?

I then have to find where the gradiet is zero, This is where my problem lies, I can see on the graph roughly where it will be, however i need the exact answer. All the books and example i have been able to find just show how to find the gradient at a specific point not a specific gradient.

I know that i haveto differentiate to get the gradient, but after that i am a little bit stuck.
Solve for x in $f~'(x)=0$, that's all it means (when there are several variables, it's another story)
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$f~'(x)=3x^2+4x-1$

Solve for x in $3x^2+4x-1=0$

You can use the discriminant or you can complete the square in order to find the solutions.

3. What i did to start the quesion was make a table of the values of X. which went up in 0.5 incrimants, then put the value of x^3, then 2x^2 then x then-2 which then gave me my values of Y.

Is that what you meant?

Using the disriminant or completeting the square, will that not just give me the roots of the derivative?

4. Originally Posted by ally79
What i did to start the quesion was make a table of the values of X. which went up in 0.5 incrimants, then put the value of x^3, then 2x^2 then x then-2 which then gave me my values of Y.

Is that what you meant?
Forget this sentence :
I think you have to solve them algebraically ^^'
That is f(0)=... ? and f(x)=0 --> x=... ?

Using the disriminant or completeting the square, will that not just give me the roots of the derivative?
Isn't it what you want :
I then have to find where the gradient is zero
?

After you find the points where the gradient is 0, find the signs of the derivative.

5. You have to forgive me i'm new to calculus, i thought they were different things, if the roots are the gradient then so be it

6. Originally Posted by ally79
You have to forgive me i'm new to calculus, i thought they were different things, if the roots are the gradient then so be it
The gradient here is like the derivative ! It's just that we rather talk about the "gradient at a point" and "the derivative of a function".
So if you want to find the points whose gradient (or derivative number) is 0, you have to find these roots

7. oki doki, I'll work it through and post the answer that i get. If you dont mind having a look at it later?