Section 4.7 : The Mean Value Theorem
5. Suppose we know that \(f\left( x \right)\) is continuous and differentiable on the interval \(\left[ { - 7,0} \right]\), that \(f\left( { - 7} \right) = - 3\) and that \(f'\left( x \right) \le 2\). What is the largest possible value for \(f\left( 0 \right)\)?
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Start SolutionWe were told in the problem statement that the function (whatever it is) satisfies the conditions of the Mean Value Theorem so let’s start out this that and plug in the known values.
\[f\left( 0 \right) - f\left( { - 7} \right) = f'\left( c \right)\left( {0 - (-7)} \right)\hspace{0.5in} \to \hspace{0.5in}f\left( 0 \right) + 3 = 7f'\left( c \right)\] Show Step 2Next, let’s solve for \(f\left( 0 \right)\).
\[f\left( 0 \right) = 7f'\left( c \right) - 3\] Show Step 3Finally, let’s take care of what we know about the derivative. We are told that the maximum value of the derivative is 2. So, plugging the maximum possible value of the derivative into \(f'\left( c \right)\) above will, in this case, give us the maximum value of \(f\left( 0 \right)\). Doing this gives,
\[f\left( 0 \right) = 7f'\left( c \right) - 3 \le 7\left( 2 \right) - 3 = 11\]So, the largest possible value for \(f\left( 0 \right)\) is 11. Or, written as an inequality this would be written as,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{f\left( 0 \right) \le 11}}\]