Question

let p and q be real numbers and let f be the function defined by: f(x)= 1+2p(x-1)+(x-1)^2, for x is greater than or equal to1 qx+p, for x>1... a. find the values of q in terms of p for which f is continuous at x=1 b.find values of p and q for which f is differentiable at x=1 c. if p and q have the values determined in part (b) id f '' a continuous function?

Answer 1

Answer on Question #82172 – Math – Calculus

Let $p$ and $q$ be real numbers and let $f$ be the function defined by:

$$f(x) = \begin{cases} 1 + 2p(x - 1) + (x - 1)^2, & x \leq 1 \\ qx + p, & x > 1 \end{cases}$$

Question

a. Find the values of $q$ in terms of $p$ for which $f$ is continuous at $x = 1$.

Solution

$$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (1 + 2p(x - 1) + (x - 1)^2) = 1$$ $$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (qx + p) = q + p$$

If $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)$, then $\lim_{x \to 1^-} f(x)$ exists and

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

Hence,

$$q + p = 1$$ $$\lim_{x \to 1^-} f(x) = 1$$ $$f(1) = 1 + 2p(1 - 1) + (1 - 1)^2 = 1 = \lim_{x \to 1^-} f(x)$$

The function $f$ is continuous at $x = 1$ if $q = 1 - p$.

Answer: the function $f$ is continuous at $x = 1$ if $q = 1 - p$.

Question

b. Find the values of $q$ in terms of $p$ for which $f$ is differentiable at $x = 1$.

Solution

$$\begin{aligned} \lim_{h \to 0^-} \frac{f(1 + h) - f(1)}{h} &= \\ &= \lim_{h \to 0^-} \frac{1 + 2p(1 + h - 1) + (1 + h - 1)^2 - (1 + 2p(1 - 1) + (1 - 1)^2)}{h} = \\ &= \lim_{h \to 0^-} \frac{1 + 2ph + h^2 - 1}{h} = \lim_{h \to 0^-} (2p + h) = 2p \end{aligned}$$ $$\begin{aligned} \lim_{h \to 0^+} \frac{f(1 + h) - f(1)}{h} &= \lim_{h \to 0^-} \frac{q(1 + h) + p - 1}{h} = \\ &= \lim_{h \to 0^-} \frac{q + p - 1}{h} + q \end{aligned}$$

If $f$ is differentiable at $x = 1$, then

$$\lim_{h \to 0^-} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0^+} \frac{f(1 + h) - f(1)}{h}$$ $$2p = \lim_{h \to 0^{-}} \frac{q + p - 1}{h} + 1$$ $$\left\{ \begin{array}{l} q + p - 1 = 0 \\ q = 2p \end{array} \right. \quad \Rightarrow \quad \left\{ \begin{array}{l} 2p + p = 1 \\ q = 2p \end{array} \right. \quad \Rightarrow \quad \left\{ \begin{array}{l} p = \frac{1}{3} \\ q = \frac{2}{3} \end{array} \right.$$

Answer: $q = \frac{2}{3}, p = \frac{1}{3}$.

Question

c. If $p$ and $q$ have the values determined in part (b) is $f''$ a continuous function?

Solution

$$\left\{ \begin{array}{l} p = \frac{1}{3} \\ q = \frac{2}{3} \end{array} \right.$$ $$f(x) = \left\{ \begin{array}{ll} 1 + \frac{2}{3}(x - 1) + (x - 1)^2, & x \leq 1 \\ \frac{2}{3}x + \frac{1}{3}, & x > 1 \end{array} \right.$$ $$f'(x) = \left\{ \begin{array}{ll} \frac{2}{3} + 2(x - 1), & x \leq 1 \\ \frac{2}{3}, & x > 1 \end{array} \right.$$ $$f'(x) = \left\{ \begin{array}{ll} 2x - \frac{4}{3}, & x \leq 1 \\ \frac{2}{3}, & x > 1 \end{array} \right.$$ $$f''(x) = 2, x < 1$$ $$f''(x) = 0, x > 1$$ $$\lim_{h \to 0^{-}} \frac{f'(1 + h) - f'(1)}{h} = \lim_{h \to 0^{-}} \frac{2(1 + h) - \frac{4}{3} - \left(2(1) - \frac{4}{3}\right)}{h} = 2$$ $$\lim_{h \to 0^{+}} \frac{f'(1 + h) - f'(1)}{h} = \lim_{h \to 0^{+}} \frac{\frac{2}{3} - \left(2(1) - \frac{4}{3}\right)}{h} = 0$$

Since $\lim_{h \to 0^{-}} \frac{f'(1 + h) - f'(1)}{h} = 2 \neq 0 = \lim_{h \to 0^{+}} \frac{f'(1 + h) - f'(1)}{h}$, then

$$\lim_{h \to 0} \frac{f'(1 + h) - f'(1)}{h} \text{ does not exist}$$

The function $f'(x)$ is not differentiable at $x = 1$.

$$f''(x) = \begin{cases} 2, & x < 1 \\ 0, & x > 1 \end{cases}$$

Therefore, $f''(x)$ is not continuous at $x = 1$.

**Answer:** $f''(x)$ is not continuous at $x = 1$.

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