Question

How do you do this calculus 1 question?

Find the linearization L(x) of the function g(x)=xf(x^2) at x=2 given the following information.
f(2)=1 f′(2)=10 f(4)=5 f′(4)=−3

Answer 1

`L(x)=10-19(x-2)`

Explanation:

To find L(x) for a function `g(x)` at a point `a` you use this equation:
`L(x)=g(a)+g'(a)(x-a)`
The point we want is `a=2` so we need to find `g(2)` and `g'(2)`.
`g(2)=2*f(2^2)`
`g(2)=2*f(4)`
because `f(4)=5`,
`g(2)=2*5`
`g(2)=10`

To fill in the last number, we must find `g'(x)`
The product rule says if `h(x)=g(x)\cdotf(x)` then `h'(x)=g'(x)\cdotf(x)+f'(x)\cdotg(x)`, so:
`g'(x)=x\cdotd/dx[f(x^2)]+1\cdotf(x^2)`
The chain rule says the derivative of a function, `f(u)`, is: `f'(u)*(du)/dx`, where `u` can be some other function. Using this, we can remove the `d/dx` in the above problem:
`g'(x)=x*f'(x^2)*d/dxx^2+f(x^2)`
Using the power rule and then simplifying:
`g'(x)=x*f'(x^2)*2x+f(x^2)`
`g'(x)=2x^2*f'(x^2)+f(x^2)`

Next, plug in `x=2`:
`g'(2)=2(2)^2*f'(2^2)+f(2^2)`
`g'(2)=8*f'(4)+f(4)`
`f'(4)=-3` and `f(4)=5`, so:
`g'(2)=8*(-3)+5`
`g'(2)=-19`

Next add everything back into the original equation:
`L(x)=10-19(x-2)`