tangent line question

[sigma]

Find a function f such that \(\displaystyle f'(x) = x^3\) and the line x+y = 0 is tangent to the graph of f.

Not quite sure where to start with this question. Could somebody point me in the right direction? Thanks.

 

[skeeter]

the line x + y = 0, which is the same as y = -x has slope = -1, so f'(x) = -1 at the point of tangency.

f'(x) = x^3

x^3 = -1

x = -1

since y = -x, y = 1

so, the point of tangency is (-1,1) which is common to both the curve and the tangent line.

now you have f'(x) and an initial condition ... find f(x).

 

[sigma]

Is this right?
\(\displaystyle \
\L\
\int {x^3 } dx = \frac{{x^4 }}{4}
\\)

but now that doesn't seem right because what am I suppose to do with (-1,1)? If I'm suppose to find f(x) from f'(x), obviously you have to do the integral of f'(x) but now I'm stuck again.

 

[daon]

Is this right?
\(\displaystyle \
\L\
\int {x^3 } dx = \frac{{x^4 }}{4}
\\)

NO! You forgot one of the most important rules! The integral always comes with a constant. My teacher would take off half credit for forgetting the +C.

 

[sigma]

Ok.\(\displaystyle \L\int {x^3 } dx = \frac{{x^4 }}{4}+ C\)

But now what? I still don't know how to finish this question. When faced with a question like this, taking the anti-derivitive would have never come to mind. Usually I do the questions like this that just involve taking the derivitive and finding the slope then putting that into your line equation but this question is totally new to me.

 

[Gene]

You have the point (-1,1) so
1=(-1)^4/4 + C
Solve for C

 

[sigma]

Thank allot. Would have never thought of that.