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Find a point c satisfying the conclusion of the Mean Value Theorem for the following function and interval. f(x)=x^(-2) [1,5]
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I know to use f(b)-f(a)/b-a becomes (1/5-1/1)/(4)
which equals -16/5
im lost after this
\[f(5)=5^{-2}=\frac{1}{25}\]
you need to solve \[\frac{f(5)-f(1)}{5-1}=f'(c)\] for \(c\)
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still lost
find dy/dx ... put it to the value calculated from above and solve for x
dy/dx = ((1/25) - 1)/(5-1) ... solve for x
2.0374 (( [1, 5]
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\[ \frac{f[5]-1}{5-1}= f'[c] \] \[ -\frac{6}{25}= -\frac 2 {c^3} \] \[3 c^3= 25 \] \[c = (\frac {25}3)^{1/3}\]
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