Question -
Answer -
(i) 
Consider
. Let x = 25 and Δx =0.3. Then,
Now, dy is approximatelyequal to Δy and is given by,
Hence, the approximatevalue of
is 0.03 + 5 = 5.03.
(ii) 
Consider. Let x = 49 and Δx =0.5. Then,
Now, dy is approximatelyequal to Δy and is given by,
Hence, the approximatevalue of
is 7 + 0.035 = 7.035.
(iii) 
Consider. Let x = 1 and Δx =− 0.4. Then,
Now, dy is approximatelyequal to Δy and is given by,
Hence, the approximatevalue of
is 1 + (−0.2) = 1 − 0.2 = 0.8.
(iv) 
Consider
. Let x = 0.008 and Δx =0.001. Then,
Now, dy is approximatelyequal to Δy and is given by,
Hence, the approximatevalue of
is 0.2 + 0.008 = 0.208.
(v) 
Consider
. Let x = 1 and Δx =−0.001. Then,
Now, dy is approximatelyequal to Δy and is given by,
Hence, the approximatevalue of
is 1 + (−0.0001) = 0.9999.
(vi) 
Consider
. Let x = 16 and Δx =−1. Then,
Now, dy is approximatelyequal to Δy and is given by,
Hence, the approximatevalue of
is 2 + (−0.03125) = 1.96875.
(vii) 
Consider
. Let x = 27 and Δx =−1. Then,
Now, dy is approximatelyequal to Δy and is given by,
Hence, the approximatevalue of
is 3 + (−0.0370) = 2.9629.
(viii) 
Consider
. Let x = 256 and Δx =−1. Then,
Now, dy is approximatelyequal to Δy and is given by,
Hence, the approximatevalue of
is 4 + (−0.0039) = 3.9961.
(ix) 
Consider. Let x = 81 and Δx =1. Then,
Now, dy is approximatelyequal to Δy and is given by,
Hence, the approximatevalue of
is 3 + 0.009 = 3.009.
(x) 
Consider
. Let x = 400 and Δx =1. Then,
Now, dy is approximatelyequal to Δy and is given by,
Hence, the approximatevalue of
is 20 + 0.025 = 20.025.