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EXAMPLE 1 Find the linearization of the function f(x) = ?x + 4 at a = 5 and use it to approximate the numbers ?8.97 and ?9.03. Are these approximations overestimates or underestimates? SOLUTION The derivative of f(x) = (x + 4)^1/2 is f'(x) = 1/2 · (x + 4)^-.5 and so we have f(5) = 3 and f'(5) = 1.5. Putting these values into this equation, we see that the linearization is L(x) = f(5) + f'(5)(x - 5) = [ ] + [ ](x - 5) = [ ] . The corresponding linear approximation is ?x + 4 ? [ ] + x/6 (when x is near 5). In particular, we have ?8.97 ? 13/6 + [ ]/6 = [ ] (round to four decimal places) and ?9.03 ? 13/6 + [ ]/6 = [ ] (round to four decimal places). The linear approximation is illustrated in the figure to the left. We see that, indeed, the tangent line approximation is a good approximation to the given function when x is near 5. We also see that our approximations are overestimates because the tangent line lies above the curve. Of course, a calculator could give us approximations for ?8.97 and ?9.03, but the linear approximation gives an approximation over an entire interval.

          EXAMPLE 1 Find the linearization of the function f(x) = ?x + 4 at a = 5 and use it to approximate the numbers ?8.97 and ?9.03. Are these approximations overestimates or underestimates?

SOLUTION The derivative of f(x) = (x + 4)^1/2 is
f'(x) = 1/2 · (x + 4)^-.5
and so we have f(5) = 3 and f'(5) = 1.5. Putting these values into this equation, we see that the linearization is
L(x) = f(5) + f'(5)(x - 5) = [ ] + [ ](x - 5)
= [ ] .

The corresponding linear approximation is
?x + 4 ? [ ] + x/6 (when x is near 5).

In particular, we have
?8.97 ? 13/6 + [ ]/6 = [ ] (round to four decimal places)
and
?9.03 ? 13/6 + [ ]/6 = [ ] (round to four decimal places).

The linear approximation is illustrated in the figure to the left. We see that, indeed, the tangent line approximation is a good approximation to the given function when x is near 5. We also see that our approximations are overestimates because the tangent line lies above the curve.
Of course, a calculator could give us approximations for ?8.97 and ?9.03, but the linear approximation gives an approximation over an entire interval.

Show more…

EXAMPLE 1 Find the linearization of the function f(x) = ?x + 4 at a = 5 and use it to approximate the numbers ?8.97 and ?9.03. Are these approximations overestimates or underestimates?

SOLUTION The derivative of f(x) = (x + 4)^1/2 is
f'(x) = 1/2 · (x + 4)^-.5
and so we have f(5) = 3 and f'(5) = 1.5. Putting these values into this equation, we see that the linearization is
L(x) = f(5) + f'(5)(x - 5) = [ ] + [ ](x - 5)
= [ ] .

The corresponding linear approximation is
?x + 4 ? [ ] + x/6 (when x is near 5).

In particular, we have
?8.97 ? 13/6 + [ ]/6 = [ ] (round to four decimal places)
and
?9.03 ? 13/6 + [ ]/6 = [ ] (round to four decimal places).

The linear approximation is illustrated in the figure to the left. We see that, indeed, the tangent line approximation is a good approximation to the given function when x is near 5. We also see that our approximations are overestimates because the tangent line lies above the curve.
Of course, a calculator could give us approximations for ?8.97 and ?9.03, but the linear approximation gives an approximation over an entire interval.

Added by Alexandra H.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert William Semus

10/01/2022

Step 1

Step 1: Find the linearization of the function \( f(x) = \sqrt{x+4} \) at \( a = 5 \). Show more…

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EXAMPLE 1 Find the linearization of the function f(x) = ∑(x + 4) at a = 5 and use it to approximate the numbers ∑8.97 and ∑9.03. Are these approximations overestimates or underestimates? SOLUTION The derivative of f(x) = (x + 4)^1/2 is f'(x) = 1/2 * (x + 4)^-0.5. Evaluating this at x = 5, we have f(5) = 3 and f'(5) = 1/6. Putting these values into the linearization equation, we see that the linearization is L(x) = f(5) + f'(5)(x - 5). The corresponding linear approximation is ∑(x + 4) ≈ 13/6 + x/6 (when x is near 5). In particular, we have: ∑8.97 ≈ 13/6 + 3.97/6 ≈ 2.9950 (round to four decimal places) and ∑9.03 ≈ 13/6 + 4.03/6 ≈ 3.0050 (round to four decimal places). The linear approximation is illustrated in the figure to the left. We see that, indeed, the tangent line approximation is a good approximation to the given function when x is near 5. We also see that our approximations are overestimates because the tangent line lies above the curve. Of course, a calculator could give us approximations for ∑8.97 and ∑9.03, but the linear approximation gives an approximation over an entire interval.

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00:01 So in this question, we are finding the linearization of the function, f of x equals the square root of x plus 4.

00:08 It equals 5, and we are using it to approximate the numbers, squared of 8 .97, and squared of 9 .03.

00:16 And we want to know, are these approximations, overestimates or underestimates? you're indeed correct that the derivative of f of x equals x plus 4 to the 1⁄2 power is 1⁄2.

00:30 Times the quantity of x plus 4.

00:34 Now, it is to the negative 0 .5, but it will be easier for us to write that as to the negative 1 half power.

00:42 And so we do have the f of 5 is equal to 3.

00:46 If i want f prime of 5, i want to clean up my f prime of x.

00:53 So if i have a negative exponent, i'm going to be flipping that into the denominator where the exponent becomes positive.

01:01 1 over 2 times x plus 4, that quantity to the 1 half power.

01:07 Now, if i want f prime of 5, i plug in.

01:13 And so i have 1 over 2 times 9 to the 1 half power.

01:20 Now, raising something to the 1 half power is the same as taking the square root of it.

01:26 The square root of 9 is 3.

01:29 So i'm getting 1 over 2 times 3, or i'm getting 1 .7.

01:33 And so my f prime of 5 this time is not 1 .5, but indeed it is one -sixth.

01:44 If i put these values into my equation, i see that my linearization, l of x, is f -of -5 plus f -prime of 5 times the quantity of x -minus 5.

01:55 It is 3 plus my f prime of 5 is 1 6th times the quantity of x minus 5.

02:08 So i could say here i have 3 plus 1 6th times the quantity of x minus 5.

02:18 Now here they might want you to simplify.

02:22 So if i had 3 plus 1 6th times the quantity of x minus 5...

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