find all the local maxima and minima (the y-values) and the values of x where they occur.
g(x)=x^3-4x+1
g'=3x^2-4=0
x= +- sqrt (3/4)
g''=6x at +sqrt (3/4) it is +, and at - sqrt (3/4) it is negative.
so what is the local maxima and minima
Well, this means you don't know any calculus.
So graph it, and read the values. Do it on your calculator.
do you have an idea how i can find it
you called it pre-calc so I guess we can't use Calculus, too bad.
Do you have use of a programmable calculator?
Some classes now use those to solve your kind of question.
You could make a table of values, graph it, then by trial and error and a scientific calculator find the largest and smallest y values you can find.
hint: the min is just a little to the left of (1,-2), and the max is around (-1,4)
yes i do the class i am in is pre calc
is the minima x=1.15 y= -2.07
maxima x=-1.15 y=4.07
Yes, you are correct to one decimal accuracy.
To find the local maxima and minima of the function g(x) = x^3 - 4x + 1, we need to first find the critical points. Critical points occur where the derivative of the function is either zero or undefined.
Step 1: Find the derivative of g(x)
g'(x) = 3x^2 - 4
Step 2: Set g'(x) equal to zero and solve for x
3x^2 - 4 = 0
To solve the quadratic equation, we can factor it or use the quadratic formula. In this case, it's easier to use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = 3, b = 0, and c = -4. Substituting these values into the quadratic formula:
x = (-0 ± √(0^2 - 4 * 3 * -4)) / (2 * 3)
x = ±√(0 + 48) / 6
x = ±√48 / 6
x = ±√8 / √6
x = ±2√2 / √6
x = ±2√2 / √6 * (√6 / √6)
x = ±2√12 / 6
x = ±2√3 / 3
So, the critical points are x = 2√3 / 3 and x = -2√3 / 3.
Step 3: To determine if these points are local maxima or minima, we can evaluate g''(x) (the second derivative) at the critical points. If g''(x) > 0, it's a local minimum. If g''(x) < 0, it's a local maximum.
Step 3a: Find the second derivative of g(x)
g''(x) = 6x
Step 3b: Evaluate the second derivative at the critical points
g''(2√3 / 3) = 6 * (2√3 / 3) = 4√3
g''(-2√3 / 3) = 6 * (-2√3 / 3) = -4√3
Since g''(2√3 / 3) > 0 and g''(-2√3 / 3) < 0, we can conclude that x = 2√3 / 3 is a local minimum, and x = -2√3 / 3 is a local maximum.
To find the corresponding y-values, we substitute the x-values into the function g(x):
g(2√3 / 3) = (2√3 / 3)^3 - 4(2√3 / 3) + 1
g(2√3 / 3) = 8√3 / 27 - 8√3 / 3 + 1
g(2√3 / 3) = (8√3 - 72√3 + 27) / 27
g(2√3 / 3) = (-64√3 + 27) / 27
g(-2√3 / 3) = (-2√3 / 3)^3 - 4(-2√3 / 3) + 1
g(-2√3 / 3) = -8√3 / 27 + 8√3 / 3 + 1
g(-2√3 / 3) = (-8√3 + 72√3 + 27) / 27
g(-2√3 / 3) = (64√3 + 27) / 27
Therefore, the local minimum y-value is (-64√3 + 27) / 27 and the local maximum y-value is (64√3 + 27) / 27.