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Let f(x)=e^(x2). By the Mean Value Theorem, there exists a point c such that


with c between 4 and 6. Approximate c to within two decimal places; we're asking you to approximate c, because you can't "solve" for c.

  • calculus -

    Your statement of
    desperately needs brackets to say
    f′(c)= ( f(6)−f(4) )/(6−4)

    f(6) = e^36 = 4.31123x10^15 , rather large
    f(4) = e^16 = only 8886110.52

    ( f(6) - f(4) )/2 = 2.1556x10^15

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