TBIL-Cal Linear Approximation (AD2)

Section 3.2 Linear Approximation (AD2)

Learning Outcomes

Use tangent lines to approximate functions.
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Subsection 3.2.1 Activities

Definition 3.2.1.

The linear approximation (or tangent line approximation or linearization) of a function \(f(x)\) at \(x=a\) is the tangent line \(L(x)\) at \(x=a\text{.}\) In formulas, \(L(x)\) is the linear function

\begin{equation*} L(x) = f'(a)(x-a) + f(a). \end{equation*}

Notice that this is obtained by writing the tangent line to \(f(x)\) at \((a,f(a))\) in point-slope form and calling the resulting linear function \(L(x)\text{.}\) The linear approximation \(L(x)\) is a linear function that looks like \(f(x)\) when we zoom in near \(x=a\text{.}\)

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Activity 3.2.2.

Without using a calculator, we will use calculus to approximate \(\ln(1.1)\text{.}\)

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(a)

Find the equation of the tangent line to \(\ln(x)\) at \(x=1\text{.}\) This will be your linear approximation \(L(x)\text{.}\) What do you get for \(L(x)\text{?}\)

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\(\displaystyle L(x) = x\)
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\(\displaystyle L(x) = x+1\)
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\(\displaystyle L(x)= x - 1\)
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\(\displaystyle L(x)= -x + 1\)
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Answer.

C: \(L(x)= x - 1\)

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(b)

As 1.1 is close to 1, we can use \(L(1.1)\) to approximate \(\ln(1.1)\text{.}\) What approximation do you get?

\(\displaystyle \ln(1.1) \approx 1.1 \)
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\(\displaystyle \ln(1.1) \approx 2.1 \)
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\(\displaystyle \ln(1.1) \approx 0.1 \)
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\(\displaystyle \ln(1.1) \approx -0.1 \)
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Answer.

C: \(\ln(1.1) \approx 0.1 \)

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(c)

Sketch the tangent line \(L(x)\) on the same plane as the graph of \(\ln(x)\text{.}\) What do you notice?

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Answer.

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Activity 3.2.3.

Using the equation of the tangent line to the graph of \(\ln(x)\) at \(x=1\) and the shape of this graph, you can show that for all values of \(x\text{,}\) we have that \(\ln(x) \leq x-1\text{.}\)

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(a)

Compute the second derivative of \(\ln(x)\text{.}\) What do you notice about the sign of the second derivative of \(\ln(x)\text{?}\) What does this tell you about the shape of the graph?

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Answer.

The second derivative is given by \(-\dfrac{1}{x^2}\text{.}\) The sign is always negative, so the shape of the graph is concave down.

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(b)

Conclude that because the graph of \(\ln(x)\) has a certain shape, the graph will bend below the tangent line and so that \(\ln(x)\) will always be smaller than the tangent line approximation \(L(x)= x -1 \text{.}\)

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Activity 3.2.4.

In this activity you will approximate power functions near \(x=1\text{.}\)

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(a)

Find the tangent line approximation to \(x^2\) at \(x=1\text{.}\)

\(\displaystyle L(x) = 2x\)
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\(\displaystyle L(x) = 2x+1\)
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\(\displaystyle L(x)= 2x - 1\)
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\(\displaystyle L(x)= -2x + 1\)
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Answer.

C: \(L(x)= 2x - 1\)

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(b)

Show that for any constant \(k\text{,}\) the tangent line approximation to \(x^k\) at \(x=1\) is \(L(x) = k(x -1) + 1\text{.}\)

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Answer.

If \(f(x) = x^k\text{,}\) then \(f'(x) = kx^{k-1}\text{.}\) So, \(f'(1) = k\text{,}\) and we have

\begin{equation*} y - 1 = k(x-1), \end{equation*}

which gives

\begin{equation*} L(x) = k(x-1) + 1. \end{equation*}

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(c)

Someone claims that the square root of 1.1 is about 1.05. Use the linear approximation to check this estimate. Do you think this estimate is about right? Why or why not?

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Answer.

The linearization is given by \(L(x) = \dfrac{1}{2}(x-1) + 1\text{,}\) so \(L(1.1) = \dfrac{1}{2}(1.1-1)+1 = 1.05\text{.}\) The estimate should be approximately correct, since 1.1 is close to 1.

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(d)

Is the actual value \(\sqrt{1.1}\) above or below 1.05? What feature of the graph of \(\sqrt{x}\) makes this an over or under estimate?

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Answer.

The actual value is 1.0488, so the approximation is an overestimate. This is an overestimate because the function bends downward, away from its tangent line.

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Remark 3.2.5.

If a function \(f(x)\) is concave up around \(x=a\text{,}\) then the function is turning upwards from its tangent line. So when we use a linear approximation, the value of the approximation will be below the actual value of the function and the approximation is an underestimate. If a function \(f(x)\) is concave down around \(x=a\text{,}\) then the function is turning downwards from its tangent line. So when we use a linear approximation, the value of the approximation will be above the actual value of the function and the approximation is an overestimate.

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Activity 3.2.6.

Suppose \(f\) has a continuous positive second derivative and \(\Delta x\) is a small increment in \(x\) (like \(h\) in the limit definition of the derivative). Which one is larger...

\begin{equation*} f(1 + \Delta x ) \quad \text{or} \quad f'(1) \Delta x + f(1) \quad? \end{equation*}

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Answer.

The function is concave up, so \(f(1+\Delta x)\) is larger.

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Activity 3.2.7.

A certain function \(p(x)\) satisfies \(p(7) = 49\) and \(p'(7) = 8\text{.}\)

Explain how to find the local linearization \(L(x)\) of \(p(x)\) at \(7\text{.}\)
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Explain how to estimate the value of \(p(6.951)\text{.}\)
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Suppose that \(p'(7)=0\) and you know that \(p''(x) < 0\) for \(x < 7\text{.}\) Explain how to determine if your estimate of \(p(6.951)\) is too large or too small.
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Suppose that \(p''(x) > 0\) for \(x > 7\text{.}\) Use this fact and the additional information above to sketch an accurate graph of \(y=p(x)\) near \(x=7\text{.}\)
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Answer.

\(\displaystyle L(x)=8 \, x - 7\)
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\(\displaystyle p(6.951) \approx 48.6064\)
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The estimate is too large.
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Activity 3.2.8.

Let’s find the quadratic polynomial

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\begin{equation*} q(x) = ax^2 +bx +c \end{equation*}

where \(a,b,c\) are parameters to be determined so that \(q(x)\) best approximates the graph of \(f(x)=\ln(x)\) at \(x=1\text{.}\)

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(a)

We want to choose \(a,b,c\) such that our quadratic polynomial resembles \(f(x)\) at \(x=1\text{.}\) First thing, we want \(f(1)=q(1)\text{.}\) What equation in \(a,b,c\) does this condition give you?

\(\displaystyle a+b+c = 1\)
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\(\displaystyle a+b+c = 0\)
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\(\displaystyle c = 0\)
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\(\displaystyle c = 1\)
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Answer.

B: \(a+b+c = 0\)

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(b)

We also want \(f'(1)=q'(1)\text{.}\) What equation in \(a,b,c\) does this condition give you?

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Answer.

\(2a + b = 1\)

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(c)

Finally, we want \(f''(1)=q''(1)\text{.}\) What equation in \(a,b,c\) does this condition give you?

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Answer.

\(2a = -1\)

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(d)

Find a solution to this system of linear equations! Your answer will give you values of \(a,b,c\) that can be used to draw a quadratic approximating the natural logarithm. You can check your answer on Desmos https://www.desmos.com/calculator/bad2xrwmvl

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Answer.

\(a = -\dfrac{1}{2}, b = 2, c= -\dfrac{3}{2}\)

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Observation 3.2.9.

A linear approximation \(L(x)\) to \(f(x)\) at \(x=a\) is a linear function with

\begin{equation*} L(a) = f(a), \quad L'(a) = f'(a). \end{equation*}

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A quadratic approximation \(Q(x)\) to \(f(x)\) at \(x=a\) is a quadratic function with

\begin{equation*} Q(a) = f(a), \quad Q'(a) = f'(a), \quad Q''(a) = f''(a) . \end{equation*}

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Activity 3.2.10.

Find the linear approximation \(L(x)\) of \(\cos(x)\) at \(x=0\text{.}\) Then find the quadratic approximation \(Q(x)\) of \(\cos(x)\) at \(x=0\text{.}\) Graph both and compare the two approximations!

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Answer.

\(L(x) = 1\) and \(Q(x) = 1- \dfrac{1}{2}x^2\)

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