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Let $f$ be a function defined by
$$f(x) = \begin{cases} x^2 &\text{ for }x \leq 1\\ ax^2+bx+c &\text{ for } 1 < x \leq 2 \\ x+d  &\text{ for } x>2 \end{cases}$$
Find the values for the constants $a$, $b$, $c$ and $d$ so that $f$ is continuous and differentiable everywhere on the real line.
asked in Calculus by Veteran (67.5k points) | 467 views

1 Answer

+9 votes
Best answer

f is differentiable at 1 if

\lim_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h}=\lim_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h}

=> 2 = 2a+b - (1)

f is differentiable at 2 if

\lim_{h\rightarrow 0^-}\frac{f(2+h)-f(2)}{h}=\lim_{h\rightarrow 0^+}\frac{f(2+h)-f(2)}{h}

=> 4a+b = 1 - (2)

Solving (1) and (2), we get

a = -0.5, b = 3

Now f has to be continous on 1 also, so

\lim_{x\rightarrow 1^-}f(x)=\lim_{x\rightarrow 1^+}\f(x) = f(1)

=> 1 = a + b + c

=> c = -1.5

Similarly f has to be continous on 2 also, so

\lim_{x\rightarrow 2^-}f(x)=\lim_{x\rightarrow 2^+}\f(x) = f(2)

=> 4a+2b+c = 2+d

=> d = 0.5

So a = -0.5, b = 3, c = -1.5, d = 0.5

answered by Veteran (11k points)
selected by
how eqn 2 comes??
Derivative when x is slightly less than 2 is 2ax+b. Putting x=2 gives you 4a+b.
how to calculate f(2+h) ?in first equation
@dq Find $f'(x)$ and directly put value $x = 2$, as we already know it is continuous and diffrentiable.


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