# How do you find the values of c and d that make the following function continuous for all x given f(x) = 9x if x<1, f(x) = cx^2+d if 1<=x<2 and f(x) = 3x if x>=2?

Jan 9, 2017

$c = - 1 , \mathmr{and} , d = 10$.

#### Explanation:

Let us name the Intervals

$x < 1 \text{ as "I_1, 1lexlt2" as "I_2," and, "xge2" as } {I}_{3}$.

On these Intervals $f$ is defined as polynomials, which, we know,

are continuous on these intervals.

So, if $f$ has to be made continuous over the whole of $\mathbb{R}$, it

has to be continuous at the joining points of these intervals; i.e., to

say, it must be continuous at $x = 1 , \mathmr{and} , x = 2$.

Now, for continuity at $x = 1$, we must have,

${\lim}_{x \rightarrow 1 -} f \left(x\right) = f \left(1\right) = {\lim}_{x \rightarrow 1 +} f \left(x\right) \ldots \ldots \ldots \ldots \ldots \left(1\right)$.

$\text{As } x \rightarrow 1 - , x < 1 \Rightarrow f \left(x\right) = 9 x \rightarrow 9.$

$\therefore {\lim}_{x \rightarrow 1 -} f \left(x\right) = 9. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(1. i\right)$

On the similar lines, we have,

${\lim}_{x \rightarrow 1 +} f \left(x\right) = c + d = f \left(1\right) \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(1. i i\right)$

$\therefore \text{ From } \left(1\right) , \left(1. i\right) \mathmr{and} , \left(1. i i\right) , c + d = 9. \ldots \ldots \ldots \ldots \ldots \ldots . . \left(I\right)$

Considering the continuity of $f \text{ at } x = 2$, we get,

$4 c + d = 6. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(I I\right)$

Solving $\left(I\right) \text{ and "(II)," we get, } c = - 1 , \mathmr{and} , d = 10$.

Enjoy Maths.!