Question

suppose that ( f(800)=3000 ) and ( f^{prime}(800)=10 ). estimate each of the following. (a) ( f(801) ) (b) ( f(800.5) ) (c) ( f(799) ) (d) ( f(798) ) (e) ( f(799.75) ) (a) ( f(801) approx 3010 ) (simplify your answer. type an exact answer.) (b) ( f(800.5) approx 3005 ) (simplify your answer. type an exact answer.) (c) ( f(799) approx 2990 ) (simplify your answer. type an exact answer.) (d) ( f(798) approx 2980 ) (simplify your answer. type an exact answer.) (e) ( f(799.75) approx 2997.5 ) (simplify your answer. type an exact answer.)

Explanation:

Step1: Use the linear approximation formula

The linear approximation formula is \(f(x)\approx f(a)+f^{\prime}(a)(x - a)\), where \(a = 800\), \(f(a)=3000\), and \(f^{\prime}(a)=10\).

Step2: Calculate \(f(801)\)

For \(x = 801\), \(x-a=801 - 800=1\). Then \(f(801)\approx f(800)+f^{\prime}(800)(801 - 800)=3000+10\times1 = 3010\).

Step3: Calculate \(f(800.5)\)

For \(x = 800.5\), \(x - a=800.5-800 = 0.5\). Then \(f(800.5)\approx f(800)+f^{\prime}(800)(800.5 - 800)=3000+10\times0.5=3005\).

Step4: Calculate \(f(799)\)

For \(x = 799\), \(x - a=799 - 800=-1\). Then \(f(799)\approx f(800)+f^{\prime}(800)(799 - 800)=3000+10\times(-1)=2990\).

Step5: Calculate \(f(798)\)

For \(x = 798\), \(x - a=798 - 800=-2\). Then \(f(798)\approx f(800)+f^{\prime}(800)(798 - 800)=3000+10\times(-2)=2980\).

Step6: Calculate \(f(799.75)\)

For \(x = 799.75\), \(x - a=799.75 - 800=-0.25\). Then \(f(799.75)\approx f(800)+f^{\prime}(800)(799.75 - 800)=3000+10\times(-0.25)=2997.5\).

Answer:

(a) \(3010\)
(b) \(3005\)
(c) \(2990\)
(d) \(2980\)
(e) \(2997.5\)