f(x)=x2−4x−3 is continuous on [1,4] as polynomial.
f(x)=x2−4x−3 is differentiable on (1,4) as polynomial.
Therefore, by the Mean Value Theorem, there is a number c in (1,4) such that
f(4)−f(1)=f′(c)(4−3)
f(4)=42−4(4)−3=−3
f(1)=12−4(1)−3=−6
f′(x)=2x−4
f′(c)=2c−4 Substitute
−3−(−6)=(2c−4)(4−1)
2c−4=1
c=25
1<25<4=>c∈(1,4) Hence the hypotheses of the Mean Value Theorem are satisfied.
Comments
Leave a comment