MHB Find the average or mean slope of the function on a interval

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To find the average slope of the function f(x)=1−8x² on the interval [-5, 6], the correct formula is [f(6) - f(-5)] / [6 - (-5)], which calculates to -8, not 8 as initially stated. The confusion arose from a miscalculation in evaluating the function at the endpoints. The derivative f'(x) = -16x was correctly derived, but setting it equal to 8 to find c was based on the incorrect average slope. Therefore, the mistake lies in the calculation of the average slope itself.
jose1
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Hello
I have the exercise below:
Consider the function f(x)=1−8x2 on the interval [−5,6]. Find the average or mean slope of the function on this interval, i.e.

[f(6)-f(-5)]/[6-(-5)]

according to the theorem of laGrange

the slope in a continues function which is derivable in an interval should equal to f'(c), So

[f(6)-f(-5)]/[6-(-5)] is equal to 8. According to the site this is wrong?

and

f'(x) = -16x. So, -16x=8 is x=-1/2. According to the site this is wrong?

Could you please give me some advice? where is my mistake?
I will appreciate any help

Thanks
 
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jose said:
Hello
I have the exercise below:
Consider the function f(x)=1−8x2 on the interval [−5,6]. Find the average or mean slope of the function on this interval, i.e.

[f(6)-f(-5)]/[6-(-5)]

according to the theorem of laGrange

the slope in a continues function which is derivable in an interval should equal to f'(c), So

[f(6)-f(-5)]/[6-(-5)] is equal to 8. According to the site this is wrong?

and

f'(x) = -16x. So, -16x=8 is x=-1/2. According to the site this is wrong?

Could you please give me some advice? where is my mistake?
I will appreciate any help

Thanks
Hello jose, and welcome to MHB.

The answer to the question is $\dfrac {f(6) - f(-5)}{6 - (-5)}$. If you check your calculation for that, you should find that it gives the value $-8$, not $8$.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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