Let's draw the tangent line at *x* = 1, since 1 is a convenient number that's close to .9. Then we have Δ *x* = -.1 *y*_{old} = *f* (1) = 1 + 3 - 2 - 1 = 1
To find the slope we need to know *f* ': *f*'(*x*) = 5*x*^{4} + 12*x*^{3 }- 4*x* - 1
The slope is *f* '(1) = 5 + 12 - 4 - 1 = 12.
This is enough to find the approximation: It's kind of weird that we got a negative number, but if we look at a graph of the function and tangent line together we can see why: the tangent line drops below the *x*-axis while the function is still above the *x*-axis. You know how to do tangent line approximations. You know how to do tangent line approximations by finding just the slope of the tangent line rather than bothering with the whole tangent line equation. The only thing we've changed for these problems is that we're no longer telling you where to draw the tangent line. Instead, you get to figure that out for yourself. To pick the right spot for the tangent line, take *x* = *a* near where you're asked to estimate *f*, and make sure *f* (*a*) and *f* '(*a*) are easy to compute. |