Finding the Tangent Line of x^4+2x^2 at x=1

[Mindscrape]

Homework Statement

Okay, so I was helping someone study for the AP calculus exam, and I don't really know why we got this question wrong. It says to find the tangent line to the equation $f(x)=x^4+2x^2$ at x=1.

Homework Equations

Taylor series, we'll just use first order.
$$f(x)|_{x=x_0} \approx f(x_0) + f'(x_0)(x-x_0)$$

The Attempt at a Solution

Should be easy as cake, but according to the AP exam answer guide the answer I got is wrong. Anyways, here's what I did.

$$f'(x)=4x^3 + 4x$$
$$f(1)=3$$
$$f'(1)=8$$

So, using our first order expansion (this level of calculus would simply call it "linearization") we should get:
$$f(x)|_{x=1} = 3 + 8x - 8 = 8x -5$$

This was one of the answers, just not the right one according to the answer sheet. The right answer was something like x - .329

I just don't see either why my answer is incorrect, and why the other answer would be correct. Am I being stupid?

 

[armolinasf]

I don't think you're being stupid. I got the same answer and I am at a loss as to how they got a y intercept of -.329 if the slope is x. I wouldn't be surprised if it was a mistake

 

[SammyS]

Homework Statement

Okay, so I was helping someone study for the AP calculus exam, and I don't really know why we got this question wrong. It says to find the tangent line to the equation $f(x)=x^4+2x^2$ at x=1.

Homework Equations

Taylor series, we'll just use first order.
$$f'(x)|_{x=x_0} \approx f(x_0) + f'(x_0)(x-x_0)$$

The Attempt at a Solution

Should be easy as cake, but according to the AP exam answer guide the answer I got is wrong. Anyways, here's what I did.

$$f'(x)=4x^3 + 4x$$
$$f(1)=3$$
$$f'(1)=8$$

So, using our first order expansion (this level of calculus would simply call it "linearization") we should get:
$$f'(x)|_{x=1} = 3 + 8x - 8 = 8x -5$$

This was one of the answers, just not the right one according to the answer sheet. The right answer was something like x - .329

I just don't see either why my answer is incorrect, and why the other answer would be correct. Am I being stupid?

Taylor series, we'll just use first order (Should be)
$$f(x) \approx f(x_0) + f'(x_0)(x-x_0)$$, not $$f'(x)|_{x=x_0}\approx\dots$$

It should be:
$$y = 3 + 8x - 8 = 8x -5$$, not $$f'(x)$$.

I would simply find $$f'(x)|_{x=1}=8$$, and use that for the slope of the tangent line.

Then find the y intercept, b, from $$y=mx+b\ \ \to\ \ 3=8(1)+b$$.

Then plug the slope & y-intercept back into the equation for the line.

 

[╔(σ_σ)╝]

I used a different method and arrived at the same answer you did.

I believe it is an error.

 

[Mindscrape]

Taylor series, we'll just use first order (Should be)
$$f(x) \approx f(x_0) + f'(x_0)(x-x_0)$$, not $$f'(x)|_{x=x_0}\approx\dots$$

It should be:
$$y = 3 + 8x - 8 = 8x -5$$, not $$f'(x)$$.

I would simply find $$f'(x)|_{x=1}=8$$, and use that for the slope of the tangent line.

Then find the y intercept, b, from $$y=mx+b\ \ \to\ \ 3=8(1)+b$$.

Then plug the slope & y-intercept back into the equation for the line.

Sorry, whoops, I don't know why I put a prime on f(x) the first time I truncated the TS (the second time I just copied and pasted). Thanks for catching that, I'll fix it now even!

Nice to know that others got the same answer.