the mean slops is just (f(4)-f(-4))/(4+4)
Since f(x) is odd, we know that f(-4) = -f(4), so
the mean slope is just 2f(4)/8 = 2*180/8 = 45
Now, f'(x) = 9x^2-3, so you want c such that
9c^2 - 3 = 45
solve for c.
see the graphs at
https://www.wolframalpha.com/input/?i=plot+y%3D3x%5E3+%E2%88%92+3x%2C+y%3D45x+for+-4%3C%3Dx%3C%3D4
to check your answers.
Consider the function f ( x ) = 3x^3 − 3x on the interval [ − 4 , 4 ] . Find the average or mean slope of the function on this interval.
By the Mean Value Theorem, we know there exists at least one c in the open interval ( − 4 , 4 ) such that f ' ( c ) is equal to this mean slope. For this problem, there are two values of c that work.
The smaller one is
the larger one is
i just need the answer with an explanation please and thanks.
2 answers
First:
Consider the function f ( x ) = 3x^3 − 3x on the interval [ − 4 , 4 ] . Find the average or mean slope of the function on this interval.
f(4) = 3*64 - 12 = 180
f(-4) = 3*-64 +12 = -180
distance between = 4 - -4 = 8
so average slope = (180 - -180) / 8 = 360/8 = 45
Then with derivative
f'(x) = 9 x^2 - 3
where does 9 x^2-3 = 45 ?
9 x^2 = 48
x^2 = 5.33
x = + 2.31 or x = -2.31
Consider the function f ( x ) = 3x^3 − 3x on the interval [ − 4 , 4 ] . Find the average or mean slope of the function on this interval.
f(4) = 3*64 - 12 = 180
f(-4) = 3*-64 +12 = -180
distance between = 4 - -4 = 8
so average slope = (180 - -180) / 8 = 360/8 = 45
Then with derivative
f'(x) = 9 x^2 - 3
where does 9 x^2-3 = 45 ?
9 x^2 = 48
x^2 = 5.33
x = + 2.31 or x = -2.31
1 answer