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10 attempts remaining.(1 point) Use linear approximation, i.e. the tangent line, to approximate 1.9^5 as follows:Let f(x) = x^5. The equation of the tangent line to f(x) at x = 2 can be written in the form y = mx + bwhere m is:-and where b is:Using this, we find our approximation for 1.9^5 is

10 attempts remaining.(1 point) Use linear approximation, i.e. the tangent line, to-example-1
User Orna
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Given:


f(x)=x^5

Tangent of line f(x) is at x=2

Line is:


y=mx+b

Find-:

Value of b, m.

Sol:

The slope of the line tangent to


f(x)=x^5

is the derivative function.

The derivative function uses the power rule.


\begin{gathered} f(x)=x^5 \\ \\ f^(\prime)(x)=5x^4 \end{gathered}

So at x=2


\begin{gathered} f^(\prime)(x)=5x^4 \\ \\ f^(\prime)(2)=5(2)^4 \\ \\ f^(\prime)(2)=80 \end{gathered}

The function value for x=2 is:


\begin{gathered} f(x)=x^5 \\ \\ f(2)=2^5 \\ \\ f(2)=32 \end{gathered}

At x=2 coordinates is:


(x_1,y_1)=(2,32)

So the general equation of a line is:


\begin{gathered} y-y_1=m(x-x_1) \\ \\ y-32=80(x-2) \end{gathered}

Solve for m and b is:


\begin{gathered} y-32=80(x-2) \\ \\ y-32=80x-160 \\ \\ y=80x-160+32 \\ \\ y=80x-128 \end{gathered}

Compare with the general equation:


\begin{gathered} y=mx+b \\ \\ y=80x-128 \end{gathered}

So value is:


\begin{gathered} m=80 \\ \\ b=-128 \end{gathered}

So, the linear approximation of 1.9^5


\begin{gathered} f(x)=x^5 \\ \\ x=1.9 \\ \\ L(x)=80x-128 \\ \\ L(1.9)=80(1.9)-128 \\ \\ =152-128 \\ \\ =24 \end{gathered}

User Tsiro
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