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  • Find the values of p and q so that f(x) = x^2+3x+p.Is differentiable at x = 1.

Find the values of p and q so that
f(x)=\left\{\begin{array}{cl} x^{2}+3 x+p, & \text { if } x \leq 1 \\ q x+2, & \text { if } x>1 \end{array}\right.
Is differentiable at x = 1.

Answers (1)

Given that,  

f(x)=\left\{\begin{array}{cl} x^{2}+3 x+p, & \text { if } x \leq 1 \\ q x+2, & \text { if } x>1 \end{array}\right.
Is differentiable at x = 1.

We know that, f(x) is differentiable at x =1 ⇔ Lf’(1) = Rf’(1).

\\ {L f^{\prime}(1)}=\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1} \\ =\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{(1-h)-1} \\ =\lim _{h \rightarrow 0} \frac{\left[\left\{(1-h)^{2}+3(1-h)+p\right]-(1+3+p)\right]}{(1-h)-1} \quad\left(\because f(x)=x^{2}+3 x+p, \text { if } x \leq 1\right) \\ =\lim _{h \rightarrow 0} \frac{\left[\left(1+h^{2}-2 h+3-3 h+p\right)-(4+p)\right]}{-h} \\ =h m \frac{\left[h^{2}-5 h+p+4-4-p\right]}{-h} \\ = \quad \lim _{h \rightarrow 0} \frac{\left[h^{2}-5 h\right]}{-h}=\lim _{h \rightarrow 0} \frac{h(h-5)}{-h} \\ =\lim _{h \rightarrow 0} h(5-h)

\\ =5 \\ \lim _{\operatorname{Rf}^{\prime}(1)}=\lim _{x \rightarrow 1} \frac{f(x)-f(1)}{x-1} \\ =\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{(1+h)-1} \\ =\lim _{h \rightarrow 0} \frac{[(q(1+h)+2)-(q+2)]}{(1+h)-1} \quad(v f(x)=q x+2, \text { if } x>1) \\ =\lim _{h \rightarrow 0} \frac{[(q+q h+2)-(q+2)]}{h} \\ =\lim _{h \rightarrow 0} \frac{[q+q h+2-q-2]}{h} \\ =q

Since, Lf’(1) = Rf’(1)

∴ 5 = q (i)

Now, we know that if a function is differentiable at a point,it is necessarily continuous at that point.

⇒ f(x) is continuous at x = 1.

⇒ f(1-) = f(1+) = f(1)

⇒ 1+3+p = q+2 = 1+3+p

⇒ p-q = 2-4 = -2

⇒ q-p = 2

Now substituting the value of ‘q’ from (i), we get

⇒ 5-p = 2

⇒ p = 3

∴ p = 3 and q = 5

 

Posted by

infoexpert22

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