Post

Created by @nathanedwards
 at October 31st 2023, 12:40:27 pm.

Question:

Let f(x)=1+xf(x) = \sqrt{1 + x} be a function defined on the interval [0,2][0, 2].

(a) Use linear approximation to estimate the value of f(1.5)f(1.5).

(b) Determine the differential of f(x)f(x) at x=1x = 1.

Answer:

(a) We can use linear approximation to estimate the value of f(1.5)f(1.5). To do this, we'll find the equation of the tangent line to the function at x=1x = 1, and then use that line to estimate f(1.5)f(1.5).

Let's find the slope of the tangent line:

f(x)=ddx(1+x)=121+xf'(x) = \frac{d}{dx}\left(\sqrt{1 + x}\right) = \frac{1}{2\sqrt{1 + x}}

We evaluate f(1)f'(1) as follows:

f(1)=121+1=14f'(1) = \frac{1}{2\sqrt{1 + 1}} = \frac{1}{4}

Now, we have the slope and a point on the line (1,f(1))=(1,2)(1, f(1)) = (1, \sqrt{2}). We can find the equation of the tangent line using the point-slope form:

yf(1)=f(1)(x1)y - f(1) = f'(1)(x - 1)
y2=14(x1)y - \sqrt{2} = \frac{1}{4}(x - 1)

We can rearrange this equation to get:

y=14x+(214)y = \frac{1}{4}x + \left(\sqrt{2} - \frac{1}{4}\right)

Now, we can estimate f(1.5)f(1.5) by evaluating the equation of the tangent line at x=1.5x = 1.5:

f(1.5)14(1.5)+(214)f(1.5) \approx \frac{1}{4}(1.5) + \left(\sqrt{2} - \frac{1}{4}\right)
f(1.5)38+214f(1.5) \approx \frac{3}{8} + \sqrt{2} - \frac{1}{4}
f(1.5)2+18f(1.5) \approx \sqrt{2} + \frac{1}{8}

Therefore, the estimate of f(1.5)f(1.5) using linear approximation is 2+18\sqrt{2} + \frac{1}{8}.

(b) To find the differential of f(x)f(x) at x=1x = 1, we'll use the differential form:

df(x)=f(x)dxdf(x) = f'(x) \cdot dx

Here, dxdx represents a small change in xx. We'll evaluate this at x=1x = 1:

df(1)=f(1)dxdf(1) = f'(1) \cdot dx
df(1)=14dxdf(1) = \frac{1}{4} \cdot dx

So, the differential of f(x)f(x) at x=1x = 1 is 14dx\frac{1}{4}dx.