Your statement of
f′(c)=f(6)−f(4)6−4
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
Let f(x)=e^(x2). By the Mean Value Theorem, there exists a point c such that
f′(c)=f(6)−f(4)6−4
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.
1 answer
1 answer