Differentiation & Stationary Points

Having some trouble with the following two questions, I have an answer but it doesn't make sense...

A function f is defined by the formula f (x) = (x-1)^2 (x+2)

(a) Find the coordinates of the points where the curve with the equation y = f(x) crosses the x - and y-axes.

(b) Find the stationary points of this curve y = f(x).

For the derivative of (a) I got 2x^2 - 3 but I didn't know how to get the points from that... Help is much appreciated!

Re: Differentiation & Stationary Points

x-intercepts are where $\displaystyle f(x) = 0$

the y-intercept is $\displaystyle f(0)$

stationary points are where $\displaystyle f'(x) = 0$

also, check your derivative. should be ...

$\displaystyle f'(x) = 3(x^2-1)$

Re: Differentiation & Stationary Points

Hello, smr101!

You seem to be lacking in much of the basics,

. . both in Calculus **and** Algebra.

Maybe a full solution (with baby steps) will jog your memory.

Quote:

Given: .$\displaystyle y \:=\:(x-1)^2(x+2)$

(a) Find the coordinates of the points where the curve crosses the x - and y-axes.

You don't know how to find intercepts?

Let $\displaystyle x = 0:\;y \:=\:(\text{-}1)^2(2) \:=\:2$

. . The *y*-intercept is $\displaystyle (0,2).$

Let $\displaystyle y = 0\!:\;0 \:=\:(x-1)^2(x+2) \quad\Rightarrow\quad x \:=\:1,\,\text{-}2$

. . The *x*-intercepts are $\displaystyle (1,0)$ and $\displaystyle (\text{-}2,0).$

Quote:

(b) Find the stationary points of this curve.

For the derivative of (a) I got 2x^2 - 3 .This is wrong.

but I didn't know how to get the points from that. .You don't?

$\displaystyle \text{We have: }\:y \;=\;\overbrace{(x-1)^2}^{f(x)}\overbrace{(x+2)}^{g(x)}$

$\displaystyle \text{Product Rule: }\;y' \;=\;f'(x)\!\cdot\!g(x) + f(x)\!\cdot\!g'(x)$

$\displaystyle \text{Hence: }\;y' \;=\;2(x-1)\!\cdot\!(x+2) + (x-1)^2\!\cdot\!1 \;=\;2(x^2 +x-2) + x^2-2x+1$

. . . . . . . . $\displaystyle =\;2x^2 + 2x - 4 + x^2 - 2x + 1 \;=\;3x^2 - 3$

$\displaystyle \text{Then: }\:3x^2 - 3 \:=\:0 \quad\Rightarrow\quad x^2 \:=\:1 \quad\Rightarrow\quad x \:=\:\pm1$

$\displaystyle \text{When }x = 1\!:\;\;y = 0$

$\displaystyle \text{When }x = \text{-}1\!:\,y = 4$

$\displaystyle \text{The stationary points are: }\:(1,0)\text{ and }(\text{-}1,4)$

Re: Differentiation & Stationary Points

Quote:

Originally Posted by

**Soroban** Hello, smr101!

You seem to be lacking in much of the basics,

. . both in Calculus **and** Algebra.

Maybe a full solution (with baby steps) will jog your memory.

Let $\displaystyle x = 0:\;y \:=\:(\text{-}1)^2(2) \:=\:2$

. . The *y*-intercept is $\displaystyle (0,2).$

Let $\displaystyle y = 0\!:\;0 \:=\:(x-1)^2(x+2) \quad\Rightarrow\quad x \:=\:1,\,\text{-}2$

. . The *x*-intercepts are $\displaystyle (1,0)$ and $\displaystyle (\text{-}2,0).$

$\displaystyle \text{We have: }\:y \;=\;\overbrace{(x-1)^2}^{f(x)}\overbrace{(x+2)}^{g(x)}$

$\displaystyle \text{Product Rule: }\;y' \;=\;f'(x)\!\cdot\!g(x) + f(x)\!\cdot\!g'(x)$

$\displaystyle \text{Hence: }\;y' \;=\;2(x-1)\!\cdot\!(x+2) + (x-1)^2\!\cdot\!1 \;=\;2(x^2 +x-2) + x^2-2x+1$

. . . . . . . . $\displaystyle =\;2x^2 + 2x - 4 + x^2 - 2x + 1 \;=\;3x^2 - 3$

$\displaystyle \text{Then: }\:3x^2 - 3 \:=\:0 \quad\Rightarrow\quad x^2 \:=\:1 \quad\Rightarrow\quad x \:=\:\pm1$

$\displaystyle \text{When }x = 1\!:\;\;y = 0$

$\displaystyle \text{When }x = \text{-}1\!:\,y = 4$

$\displaystyle \text{The stationary points are: }\:(1,0)\text{ and }(\text{-}1,4)$

Thank you!

It isn't that I don't know how to find y/x intercepts, it's that my derivative answer didn't look right (as you said, it wasn't) which threw me off.