Given the function $f (x) = 3 x^{3} - 2 x$ $f(x)=3x^3−2x$ , how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [-4,4] and find the c?

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Answer 1 · 2017-04-01T10:10:19

The values of $c$ $c$ are $= \pm 2.31$ $=+-2.31$

The mean value states theorem that if a function is continous on an

interval $[a, b]$ $[a,b]$ and derivable on the interval $] a, b [$ $]a,b[$ , then there is a

point $c \in [a, b]$ $c in [a,b]$ such that $f'(c)=(f(b)-f(a))/(b-a)$

Here,

$f(x)=3x^3-2x$ is a polynomial function, defined, continuous and derivable on the interval $x in [-4,4]$

$f'(x)=9x^2-2$

$f(-4)=3*(-4)^3+8=-192+8=-184$

$f(4)=3*4^3-8=184$

Therefore,

$f'(c)=(f(4)-f(-4))/(4+4)=(184+184)/8=46$

$f'(c)=9c^2-2=46$

$c^2=(46+2)/9=sqrt48/3$

$c=+-4sqrt3/3=+-2.31$

Therefore,

$c in [-4,4]$

Given the function f(x)=3x^3−2xf(x)=3x3−2x f(x)=3x^3−2x, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [-4,4] and find the c?