Mathematics Calculus
Find and classify all locsl extreme values of each given function.(no decimals)
(a) g(x)=e^(2x)+e^(-x)
(b) g(x)=x^(2/3)*(x^(2)-4)
You will find here at Jiskha that long series of questions, posted with no evidence of effort or thought by the person posting, will not be answered. We will gladly respond to your future questions in which your thoughts are included.
Or you can post what you've done so far for each of the questions you have posted, and a math tutor would be more likely to help.
You can ask a new question or answer this question .
Similar Questions
Top answer:
To find the local extreme values of a function, we need to find its critical points first. Critical
Read more.
Top answer:
Set the derivatives equal to zero to determine the location of the extreme values. Determine what
Read more.
Top answer:
f'(x) = (x^4 + 4x^3) e^x = x^3(x+4)e^x f'(x)=0 at x = 0, -4 so the extremes are (-4, 256/e^4), (0,0)
Read more.
For f(x)=2(x+5)^3 +7 Find and classify the extreme values, determine where the function is increasing and decreasing, where it is
Top answer:
Think of the graph for y = x^3 It has no max/min, but just an inflection point at x=0. You function
Read more.
Top answer:
Your derivative is wrong, you should have y' = x^3 - 2x^2 + x When I set x^3 - 2x^2 - x = 0, I get
Read more.
Top answer:
To find the extreme values of the function, we need to find the critical points. To do this, we take
Read more.
Top answer:
To find the values for a and b that maximize the function f(x) = ax * e^(bx) given that f(1) = 3, we
Read more.
Top answer:
I think there's posibly a piece of information missing in the question as you've stated it, which is
Read more.
Top answer:
extreme where y' = 0 decreasing where y' < 0 y' = (3x-2)(x-2) so, it should be clear where y'=0. y'
Read more.
g(x)=(x^3)-(3x^2)+17 a)find and classify all critical points of g(x) using exact with x and y values. b)for what values of x is
Top answer:
To find and classify the critical points of the function g(x)=(x^3)-(3x^2)+17, we need to find the
Read more.