If f(1)=-2 & f'(x)_< 1.8 for 1_< x_< 8, what is the largest possible value of f(8)?
Can someone help me on this question and plz provid me some calculation for this question.
the largest value for f(8) would be where the slope is a maximum (1.8) for 1<=x<=8.
That would be the line
(y+2)/(x-1) = 1.8
y = 1.8x - 3.8
f(8) = 10.6
Thanks steve
To find the largest possible value of f(8), we'll need to analyze the given information and apply some mathematical concepts.
We know that f'(x) is less than or equal to 1.8 for 1 ≤ x ≤ 8. The derivative of a function represents its rate of change at a given point. So, if the derivative is less than or equal to a certain value over a range, it means the function is increasing but not too fast.
To determine the largest possible value of f(8), we can use the concept of integration. Since the derivative represents the rate of change, we can integrate it over the given range to obtain the original function f(x).
First, let's integrate f'(x) from 1 to 8:
∫(f'(x))dx = ∫(1.8)dx = 1.8x + C
Now, we need to determine the constant value C. We know that f(1) = -2, so let's substitute x = 1 and f(x) = -2 into our integrated equation:
1.8(1) + C = -2
1.8 + C = -2
C = -3.8
Now, the equation for f(x) is:
f(x) = 1.8x - 3.8
To find the largest possible value of f(8), we substitute x = 8 into our equation:
f(8) = 1.8(8) - 3.8
f(8) = 14.4 - 3.8
f(8) = 10.6
Therefore, the largest possible value of f(8) is 10.6.