hello all
i'm really happy caz i found a furums that will help me in my homework and this is also quaetions can help me plz/
thax 4 all agian.
i suppose these are calculus questions...
for image003 the function $\displaystyle f(x) = 3 + 4x^3 - x^4$..
a)
$\displaystyle f'(x) = 12x^2 -4x^3$
at x=1, $\displaystyle f'(x)=8$, hence the equation of the tangent line at the point (1,6) is y-6 = 8(x-1) and the equation of the line perpendicular to it is $\displaystyle y-6=\frac{-1}{8}(x-1)$
b) solve for $\displaystyle f'(x) = 0$ and plug in the values of x in f(x) to find their corresponding ordinates..
c) on the interval [-1,2]
f(-1) = -2
f(2) = 19
and f(-1)f(2) = -38 < 0
so, by a corollary to the IVT, there is a c such that f(c) = 0.
indeed, $\displaystyle f(c) = 3 + 4c^3 - c^4 = 0$, solving for c, we have (using a calculating device) -.8519, and 4.0453
Ü