Given:
[tex]f(x)=x^5[/tex]
Tangent of line f(x) is at x=2
Line is:
[tex]y=mx+b[/tex]
Find-:
Value of b, m.
Sol:
The slope of the line tangent to
[tex]f(x)=x^5[/tex]
is the derivative function.
The derivative function uses the power rule.
[tex]\begin{gathered} f(x)=x^5 \\ \\ f^{\prime}(x)=5x^4 \end{gathered}[/tex]
So at x=2
[tex]\begin{gathered} f^{\prime}(x)=5x^4 \\ \\ f^{\prime}(2)=5(2)^4 \\ \\ f^{\prime}(2)=80 \end{gathered}[/tex]
The function value for x=2 is:
[tex]\begin{gathered} f(x)=x^5 \\ \\ f(2)=2^5 \\ \\ f(2)=32 \end{gathered}[/tex]
At x=2 coordinates is:
[tex](x_1,y_1)=(2,32)[/tex]
So the general equation of a line is:
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ \\ y-32=80(x-2) \end{gathered}[/tex]
Solve for m and b is:
[tex]\begin{gathered} y-32=80(x-2) \\ \\ y-32=80x-160 \\ \\ y=80x-160+32 \\ \\ y=80x-128 \end{gathered}[/tex]
Compare with the general equation:
[tex]\begin{gathered} y=mx+b \\ \\ y=80x-128 \end{gathered}[/tex]
So value is:
[tex]\begin{gathered} m=80 \\ \\ b=-128 \end{gathered}[/tex]
So, the linear approximation of 1.9^5
[tex]\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}[/tex]
