# Using the Tangent Line to Approximate Function Values

## Key Questions

• A formula for a tangent line approximation of a function f, also called linear approximation , is given by

$f \left(x\right) \approx f \left(a\right) + f ' \left(a\right) \left(x - a\right) ,$
which is a good approximation for $x$ when it is close enough to $a$.

I'm not sure, but I think the question is about approximate the value $\ln \left(1.004\right)$. Could you verify it please? Otherwise we will need to know an approximation to $\ln \left(10\right) .$

In this case, we have $f \left(x\right) = \ln \left(x\right)$, $x = 1.004$ and $a = 1$.

Since,
$f ' \left(x\right) = \frac{1}{x} \implies f ' \left(1\right) = 1$ and $\ln \left(1\right) = 0$, we get

$\ln \left(1.004\right) \approx \ln \left(1\right) + 1 \cdot \left(1.004 - 1\right) = 0.004 .$

• We need to find the derivative of $f \left(x\right)$. We need to use the Chain Rule to find the derivative of $f \left(x\right)$.

$f \left(x\right) = \sqrt{1 + x} = {\left(1 + x\right)}^{\frac{1}{2}}$

$f ' \left(x\right) = \left(\frac{1}{2}\right) {\left(1 + x\right)}^{\left(\frac{1}{2} - 1\right)} \cdot 1$

$f ' \left(x\right) = \left(\frac{1}{2}\right) {\left(1 + x\right)}^{\left(\frac{1}{2} - \frac{2}{2}\right)}$

$f ' \left(x\right) = \left(\frac{1}{2}\right) {\left(1 + x\right)}^{\left(- \frac{1}{2}\right)}$

$f ' \left(x\right) = \frac{1}{2 \sqrt{1 + x}}$

$f ' \left(0\right) = \frac{1}{2 \sqrt{1 + 0}} = \frac{1}{2 \sqrt{1}} = \frac{1}{2} = 0.5$

• The linear approximation $L \left(x\right)$ of a function is done using the tangent line to the graph of the function. The equation of the tangent line at $x = a$ is given by

$y = f ' \left(a\right) \left(x - a\right) + f \left(a\right)$,

the linear approximation is

$L \left(x\right) = f ' \left(a\right) \left(x - a\right) + f \left(a\right)$.