Q: ) f(x) = 12x5 – 45x4 + 40x3 + 5. Find the value of x for which the curve shows relative maxima & relative minima

This is really simple if you watched videos. Find the first derivative of a function f(x) and find the critical numbers. Then, find the second derivative of a function f(x) and put the critical numbers. If the value is negative, the function has relative maxima at that point, if the value is positive, the function has relative maxima at that point. This is the Second Derivative Test. However, if you get 0, you have to use the First Derivative Test. Just find the first derivative of a function f(x) and critical numbers. Then, divide the domain (all real numbers) by the critical numbers. For example, if the critical numbers are -1, 4, 5, you should get 4 different domains which are (-∞, -1), (-1, 4), (4, 5), (5, ∞). Then, input any number inside each domain. If the value is negative and the value of next domain is positive, it has relative minima at that point and vice versa. For example, if you got a negative value in the interval (-1, 4) and positive value in the interval (4, 5), the function has relative minima at point 4. Hope this helps! If you have any questions or need help, please ask! :)

Q: given the function f(x) = x^3+ax^2+bx+12, when x=3 there is a relative minimum value of -15. solve for a and b

You know that plugging in x=3 will give a value of -15, so that gives you an equation in a and b. If you take the derivative of f and plug in x=3, you know it will give 0, since you have a relative minimum. This gives you a second equation in a and b. Solve the system for a and b.

Q: I don understand how to do it without a graph.

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

Question 1

(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

Accepted Answer

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))

Question 2

what if there are 2 variables F(x,y) = x^3 +y^2 -xy +x

Accepted Answer

Then you're doing multivariable calculus, which is covered in a different section of Khan Academy. You can find it in the Courses dropdown.

Question 3

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

Accepted Answer

You find all of the relative maxima, nondifferentiable points, and endpoints of the intervals, and see which one gives the highest function value.

Question 4

How do you solve for x^2/(x^2 -1)

Accepted Answer

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

Question 5

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.

Accepted Answer

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.

Question 6

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?

Accepted Answer

I'm pretty sure that means there are no extreme values

Question 7

How would we calculate the local minimum and maximum of a function defined as min{abs(x-1), x}?

Accepted Answer

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

Question 8

) f(x) = 12x5 – 45x4 + 40x3 + 5. Find the value of x for which the curve shows relative maxima & relative minima

Accepted Answer

This is really simple if you watched videos. Find the first derivative of a function f(x) and find the critical numbers. Then, find the second derivative of a function f(x) and put the critical numbers. If the value is negative, the function has relative maxima at that point, if the value is positive, the function has relative maxima at that point. This is the Second Derivative Test. However, if you get 0, you have to use the First Derivative Test. Just find the first derivative of a function f(x) and critical numbers. Then, divide the domain (all real numbers) by the critical numbers. For example, if the critical numbers are -1, 4, 5, you should get 4 different domains which are (-∞, -1), (-1, 4), (4, 5), (5, ∞). Then, input any number inside each domain. If the value is negative and the value of next domain is positive, it has relative minima at that point and vice versa. For example, if you got a negative value in the interval (-1, 4) and positive value in the interval (4, 5), the function has relative minima at point 4.

Hope this helps! If you have any questions or need help, please ask! :)

Hope this helps! If you have any questions or need help, please ask! :)

Question 9

given the function f(x) = x^3+ax^2+bx+12, when x=3 there is a relative minimum value of -15. solve for a and b

Accepted Answer

You know that plugging in x=3 will give a value of -15, so that gives you an equation in a and b.

If you take the derivative of f and plug in x=3, you know it will give 0, since you have a relative minimum. This gives you a second equation in a and b. Solve the system for a and b.

Question 10

I don understand how to do it without a graph.

Accepted Answer

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

Interval	$x$ ‍-value	$f^{'} (x)$ ‍	Verdict
$x < - 3$ ‍	$x = - 4$ ‍	$f^{'} (- 4) = 15 > 0$ ‍	$f$ ‍ is increasing. $↗$ ‍
$- 3 < x < 1$ ‍	$x = 0$ ‍	$f^{'} (0) = - 9 < 0$ ‍	$f$ ‍ is decreasing. $↘$ ‍
$x > 1$ ‍	$x = 2$ ‍	$f^{'} (2) = 15 > 0$ ‍	$f$ ‍ is increasing. $↗$ ‍

$x$ ‍	Before	After	Verdict
$- 3$ ‍	$↗$ ‍	$↘$ ‍	Maximum
$1$ ‍	$↘$ ‍	$↗$ ‍	Minimum

Course: Calculus, all content (2017 edition) > Unit 3

Relative minima & maxima review

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

Example

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