1. ## Functions

hi,

Q1.

f(x)=(x-2)(x+1)^2

a) find the min and max values of f(x):

b) A point of inflection is a point where the graph changes direction of concavity, i.e. a point where f''(x)=0. Find the inflection point(s) of f, and determine where the function is concave up and concave down.

ok..

for a),
i presumed that if y'=0 then it is either the min/max value,
so using product rule, i got 0=2(x+1)(x-2)
i.e. x=-1 or x=2
so with those x values, the function will be at its min = 0,
i also presumed that the max value is infinity...

for b), well im stuck on that

could anyone help me on both questions?

thanks,

moon

2. Originally Posted by Moon Hoplite
hi,

Q1.

f(x)=(x-2)(x+1)^2

a) find the min and max values of f(x):

b) A point of inflection is a point where the graph changes direction of concavity, i.e. a point where f''(x)=0. Find the inflection point(s) of f, and determine where the function is concave up and concave down.

ok..

for a),
i presumed that if y'=0 then it is either the min/max value,
so using product rule, i got 0=2(x+1)(x-2)
i.e. x=-1 or x=2
so with those x values, the function will be at its min = 0,
i also presumed that the max value is infinity...

for b), well im stuck on that

could anyone help me on both questions?

thanks,

moon
The product rule states that
$\displaystyle \frac{d}{dx} f(x) g(x) = f'(x) g(x) + g'(x) f(x)$

Thus, the correct derivative of the function $\displaystyle f(x) = (x-2)(x+1)^2$ is:
$\displaystyle (x+1)^2 + 2(x+1)(x-2) = (x+1)(x+1+2x-4) = 3(x-1)(x+1)$

Now you say $\displaystyle \frac{d}{dx} f(x) = 3(x-1)(x+1) = 0$ and solve for $\displaystyle x$, which gives you extreme points at $\displaystyle x=-1,1$.

To find the points of inflection or to determine concavity, differentiate once again: $\displaystyle \frac{d^2}{dx^2} = 3(x+1) + 3(x-1) = 6x = 0$. Your only inflection point is at $\displaystyle x=0$.

To verify whether $\displaystyle x=-1,1$ is a minimum or maximum, check the concavity at those points. So plug in $\displaystyle x=-1$ into $\displaystyle \frac{d^2}{dx^2} = 6x$ to see if it is positive or negative. If it is positive, then it is concave up (and thus a min) and negative means concave down (and thus a max). Do the same for $\displaystyle x=1$.

It helps to graph the function to verify.

3. Originally Posted by Last_Singularity
The product rule states that
$\displaystyle \frac{d}{dx} f(x) g(x) = f'(x) g(x) + g'(x) f(x)$

Thus, the correct derivative of the function $\displaystyle f(x) = (x-2)(x+1)^2$ is:
$\displaystyle (x+1)^2 + 2(x+1)(x-2) = (x+1)(x+1+2x-4) = 3(x-1)(x+1)$

Now you say $\displaystyle \frac{d}{dx} f(x) = 3(x-1)(x+1) = 0$ and solve for $\displaystyle x$, which gives you extreme points at $\displaystyle x=-1,1$.

To find the points of inflection or to determine concavity, differentiate once again: $\displaystyle \frac{d^2}{dx^2} = 3(x+1) + 3(x-1) = 6x = 0$. Your only inflection point is at $\displaystyle x=0$.

To verify whether $\displaystyle x=-1,1$ is a minimum or maximum, check the concavity at those points. So plug in $\displaystyle x=-1$ into $\displaystyle \frac{d^2}{dx^2} = 6x$ to see if it is positive or negative. If it is positive, then it is concave up (and thus a min) and negative means concave down (and thus a max). Do the same for $\displaystyle x=1$.

It helps to graph the function to verify.

hmmn thanks, but im confused on question 1, so does y'=0 states the min or max x values? and if so is the max value at infinity?

4. To verify whether $\displaystyle x=-1,1$ is a minimum or maximum, check the concavity at those points. So plug in $\displaystyle x=-1$ into $\displaystyle \frac{d^2}{dx^2} = 6x$ to see if it is positive or negative. If it is positive, then it is concave up (and thus a min) and negative means concave down (and thus a max). Do the same for $\displaystyle x=1$.

To verify whether $\displaystyle x=-1,1$ is a minimum or maximum, check the concavity at those points. So plug in $\displaystyle x=-1$ into $\displaystyle \frac{d^2}{dx^2} = 6x$ to see if it is positive or negative. If it is positive, then it is concave up (and thus a min) and negative means concave down (and thus a max). Do the same for $\displaystyle x=1$.

To verify whether $\displaystyle x=-1,1$ is a minimum or maximum, check the concavity at those points. So plug in $\displaystyle x=-1$ into $\displaystyle \frac{d^2}{dx^2} = 6x$ to see if it is positive or negative. If it is positive, then it is concave up (and thus a min) and negative means concave down (and thus a max). Do the same for $\displaystyle x=1$.

To verify whether $\displaystyle x=-1,1$ is a minimum or maximum, check the concavity at those points. So plug in $\displaystyle x=-1$ into $\displaystyle \frac{d^2}{dx^2} = 6x$ to see if it is positive or negative. If it is positive, then it is concave up (and thus a min) and negative means concave down (and thus a max). Do the same for $\displaystyle x=1$.