In the interval [0,3] f(x)= {1 , x=0
ax+b , 0 less than x less than or equal to 1
x2+4x+c , 1 less than x less than or equal to 3
(if it isnt clear, f(x) is a piecewise function)
How do i solve for a, b and c?
thanks!
Anyone?? Please??
In the interval [0,3] f(x)= {1 , x=0
ax+b , 0 less than x less than or equal to 1
x2+4x+c , 1 less than x less than or equal to 3
(if it isnt clear, f(x) is a piecewise function)
How do i solve for a, b and c?
thanks!
Anyone?? Please??
f(x) needs to be continuous on [0,3] and differentiable on (0,3).
$\displaystyle \lim_{x \to 0^+} ax + b = 1$
so , that means $\displaystyle b = 1$
to be continuous at x = 1 ...
$\displaystyle f(1) = a+1$
$\displaystyle \lim_{x \to 1^-} ax+1 = \lim_{x \to 1^+} x^2 + 4x + c$
$\displaystyle a+1 = 5+c$
to be differentiable at x = 1 ...
$\displaystyle \lim_{x \to 1^-} \frac{(ax+1) - (a+1)}{x - 1}$
$\displaystyle \lim_{x \to 1^-} \frac{a(x-1)}{x - 1} = a$
$\displaystyle \lim_{x \to 1^+} \frac{(x^2+4x+c) - (5+c)}{x - 1}$
$\displaystyle \lim_{x \to 1^+} \frac{x^2+4x-5}{x - 1}$
$\displaystyle \lim_{x \to 1^+} \frac{(x+5)(x-1)}{x - 1} = 6$
$\displaystyle a = 6$
$\displaystyle c = 2$