The function is given as
y = 7x^3 - 8x^2 -16x +15
The max is at x = -4/7 & min is at x = 4/3
How do I know that it is local max&min or absolute max &min? (graphing calculator is not allowed)
You know that it's a local minimum and maximum because the x^3 term approaches positive infinity as x goes to infinity and negative infinity as x goes to minus infinity.
explanation please?
Generally if the domain or interval is not given, how do we assume the extrema is local or absolute?
If the domain is not given then you can assume it is the maximal possible domain. In this case it is the entire real axis, but in some cases you have to exclude regions where the function is not defined.
You then determine the range of the function. To this this you need to know the properties of the function. In this case it is simple; you are dealing with a polynomial function. The x^3 term dominates for large positive and negative x.
So, you already know by just looking at the function that the range of the function is from minus infinity to plus infinity (note that (-x)^3 = - x^3 ).
This means that the points where the derivatives are zero can't be absolute maxima or minima.
If you have a polynomial of even degree, e.g. with a x^4 term then, depending on the sign of the x^4 term, the range extends either from some minumum value to plus infinity or to negative infinity, but not from minus to plus infinity.
If the range extends to plus infinity, then you can only have local maxima, but a minimum can be global. You should then list al the minima and examine at which of these the value of the function is the least.
To determine whether the extrema are local or absolute, you need to examine the behavior of the function at the critical points.
In this case, the critical points occur when the derivative of the function is equal to zero. To find these points, you can take the derivative of the function y = 7x^3 - 8x^2 - 16x + 15.
Differentiating the function, you get:
y' = 21x^2 - 16x - 16
Setting y' equal to zero, you have:
21x^2 - 16x - 16 = 0
To solve this quadratic equation, you can either factor it or use the quadratic formula.
However, before solving for x, it is important to note that if the critical points lie within the domain of the function, then they could be potential local extrema. If the critical points lie outside the domain of the function, they cannot be considered local extrema since the function does not exist at those points.
In this case, the function is defined for all real values of x, so the critical points are within the domain.
Solving the quadratic equation, you find two critical points:
x = -4/7 and x = 4/3
To determine whether these critical points are local minima or maxima, you need to examine the behavior of the function near these points.
You can do this by checking the sign of the second derivative at each critical point. The second derivative reveals information about the concavity of the function.
To find the second derivative, you need to differentiate the first derivative with respect to x.
Differentiating the first derivative, you get:
y'' = 42x - 16
Plugging in the values of x for the critical points:
For x = -4/7:
y'' = 42(-4/7) - 16 = -24 < 0
For x = 4/3:
y'' = 42(4/3) - 16 = 24 > 0
Since the second derivative is negative at x = -4/7, the point (-4/7, f(-4/7)) represents a local maximum.
Similarly, since the second derivative is positive at x = 4/3, the point (4/3, f(4/3)) represents a local minimum.
In conclusion, for this function, the points at x = -4/7 and x = 4/3 are local maxima and minima, respectively.