To get the linear approximation, we follow the equation below:
[tex]y=f(a)+f^{\prime}(a)(x-a)[/tex]where "a" is the given value of x and f'(a) is the slope of the function at a given value of "a".
In the given equation, the given value of "a" or x is 5.
Let's now solve for the linear approximation. Here are the steps:
1. Solve for f(a) by replacing the x-variable in the given function with 5.
[tex]f(5)=5^5[/tex][tex]f(a)=3125[/tex]The value of f(a) is 3125.
2. Solve for the first derivative of f(x) using the power rule.
[tex]f(x)=x^5\Rightarrow f^{\prime}(x)=5x^4[/tex]The first derivative is equal to 5x⁴.
3. Replace the "x" variable in the first derivative with 5 and solve.
[tex]f^{\prime}(5)=5(5)^4[/tex][tex]f^{\prime}(5)=5(625)[/tex][tex]f^{\prime}(5)=3125[/tex]The value of the first derivative at x = 5 is also 3,125.
4. Using the linear approximation formula above, let's now replace f(a) with 3125 and f'(a) with 3125 as well since those are the calculated value in steps 1 and 3. Replace "a' with 5 too.
[tex]y=3125+3125(x-5)[/tex][tex]y=3125+3125(x-5)[/tex]5. Simplify the equation above.
[tex]y=3125+3125x-15625[/tex][tex]y=3125x-12500[/tex]Hence, the equation of the tangent line to f(x) at x = 5 is y = 3,125x - 12500 where the slope m is 3,125 and the y-intercept b is -12,500.
Now, to find our approximation for 4.7⁵, replace the "x" variable in the equation of the tangent line with 4.7 and solve.
[tex]y=3,125x-12,500[/tex][tex]y=3,125(4.7)-12,500[/tex][tex]y=14,687.5-12,500[/tex][tex]y=2187.5[/tex]
Using the approximated linear equation, the approximated value of 4.7^5 is 2, 187.5.
