Question

I have a calculus question about linear approximation. It is a doozie. High school, 12th grade senior AP Calculus. Math, not physics.

Answer 1

To get the linear approximation, we follow the equation below:

`y=f(a)+f^(\prime)(a)(x-a)`

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.

`f(5)=5^5`
`f(a)=3125`

The value of f(a) is 3125.

2. Solve for the first derivative of f(x) using the power rule.

`f(x)=x^5\Rightarrow f^(\prime)(x)=5x^4`

The first derivative is equal to 5x⁴.

3. Replace the "x" variable in the first derivative with 5 and solve.

`f^(\prime)(5)=5(5)^4`
`f^(\prime)(5)=5(625)`
`f^(\prime)(5)=3125`

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.

`y=3125+3125(x-5)`
`y=3125+3125(x-5)`

5. Simplify the equation above.

`y=3125+3125x-15625`
`y=3125x-12500`

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.

`y=3,125x-12,500`
`y=3,125(4.7)-12,500`
`y=14,687.5-12,500`

`y=2187.5`

Using the approximated linear equation, the approximated value of 4.7^5 is 2, 187.5.