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### Course: AP®︎/College Calculus AB>Unit 5

Lesson 4: Using the first derivative test to find relative (local) extrema

# Relative minima & maxima review

Review how we use differential calculus to find relative extremum (minimum and maximum) points.

## How do I find relative minimum & maximum points with differential calculus?

A relative maximum point is a point where the function changes direction from increasing to decreasing (making that point a "peak" in the graph).
Similarly, a relative minimum point is a point where the function changes direction from decreasing to increasing (making that point a "bottom" in the graph).
Supposing you already know how to find increasing & decreasing intervals of a function, finding relative extremum points involves one more step: finding the points where the function changes direction.

## Example

Let's find the relative extremum points of $f\left(x\right)={x}^{3}+3{x}^{2}-9x+7$. First, we differentiate $f$:
${f}^{\prime }\left(x\right)=3\left(x+3\right)\left(x-1\right)$
Our critical points are $x=-3$ and $x=1$.
Let's evaluate ${f}^{\prime }$ at each interval to see if it's positive or negative on that interval.
Interval$x$-value${f}^{\prime }\left(x\right)$Verdict
$x<-3$$x=-4$${f}^{\prime }\left(-4\right)=15>0$$f$ is increasing. $↗$
$-3$x=0$${f}^{\prime }\left(0\right)=-9<0$$f$ is decreasing. $↘$
$x>1$$x=2$${f}^{\prime }\left(2\right)=15>0$$f$ is increasing. $↗$
Now let's look at the critical points:
$x$BeforeAfterVerdict
$-3$$↗$$↘$Maximum
$1$$↘$$↗$Minimum
In conclusion, the function has a maximum point at $x=-3$ and a minimum point at $x=1$.

Problem 1
$h\left(x\right)=-{x}^{3}+3{x}^{2}-4$
For what value of $x$ does $h$ have a relative maximum ?

Want to try more problems like this? Check out this exercise.

## Want to join the conversation?

• (x−3)^2(1)+(x−2)[2(x−3)]
= (x-3)^2+((x-2)2)(x-3)
= (x-3)^2+(2x-4)(x-3)
.... then I am lost. how do I get from here to ...
​=(x−3)(x−3+2x−4) ??

Thank you
• Think of this equation: x^2 + 2x. You can split the x^2 term into the individual x's and reverse the distributive property. You end up with x(x+2). The same can be done with your equation.
(x-3)^2 = (x-3)(x-3)
(x-3)(x-3) + (2x-4)(x-3)
(x-3)((x-3) + (2x-4))
• I don understand how to do it without a graph.
(1 vote)
• If you don't want to hear me take a long winded explanation of why it works then just skip.
If you think of maximum, it's like a hill. Say you can only climb the hill form the left to the right. If you're beginning to climb it, it's sloping up. But once you reach the top, it will start sloping down. And Since it must go from sloping up to sloping down in a continuous fashion the top point must have a slope of 0. (vice versa for minimum)

First you take the derivative of an arbitrary function f(x). So now you have f'(x). Find all the x values for which f'(x) = 0 and list them down. So say the function f'(x) is 0 at the points x1,x2 and x3. Now test the points in between the points and if it goes from + to 0 to - then its a maximum and if it goes from - to 0 to + its a minimum
• what if there are 2 variables F(x,y) = x^3 +y^2 -xy +x
• Then you're doing multivariable calculus, which is covered in a different section of Khan Academy. You can find it in the Courses dropdown.
• so if f'(x) goes from + to - at x=c then f(c) is an maximum but how do i know if it is a relative or absolute max
• You find all of the relative maxima, nondifferentiable points, and endpoints of the intervals, and see which one gives the highest function value.
• When we're doing the number line, what if it is negative before and after the point? Does that mean it is just continuously decreasing?
• I'm pretty sure that means there are no extreme values
• How do you solve for x^2/(x^2 -1)
• you can add and subtract 1 from numerator, so we will get:
x^2 + 1 - 1/(x^2-1) => x^2-1/x^2-1 + 1/x^2-1 => 1 + 1/x^2-1

you can solve it further given what the equation is equated to, whether it is 0 or other
• Values less than one are increasing (becoming less negative) and not decreasing, therefore there should only be one point with that is a retaliative minimum, that point being when x=1.
• Remember that the derivative is the slope at any given point. If the slope is less than one (negative) the graph is decreasing at that point.
• How would we calculate the local minimum and maximum of a function defined as min{abs(x-1), x}?
• Let 𝑓(𝑥) = min(|𝑥 − 1|, 𝑥)

What I did was to first write |𝑥 − 1| as a piecewise function:
1 − 𝑥 for 𝑥 < 1
𝑥 − 1 for 𝑥 ≥ 1

Then I found out when this function is less than 𝑥.
1 − 𝑥 < 𝑥 ⇒ 𝑥 > 1∕2
𝑥 − 1 < 𝑥 ⇒ −1 < 0, which is true for all 𝑥

That way, I could write 𝑓(𝑥) as a piecewise function:
𝑥 for 𝑥 ≤ 1∕2
1 − 𝑥 for 𝑥 ∈ (1∕2, 1)
𝑥 − 1 for 𝑥 ≥ 1

Both 𝑥 and |𝑥 − 1| are continuous and thereby 𝑓(𝑥) is also continuous.

Finally, I realized that
𝑥 has a positive slope,
1 − 𝑥 has a negative slope
and 𝑥 − 1 has a positive slope
which means that 𝑓(𝑥) has a relative maximum at 𝑥 = 1∕2
and a relative minimum at 𝑥 = 1