Mathematics, 23.07.2019 05:30, Fireburntbudder

Find the values of m and b that make the following function differentiable. the piecewise function f of x equals x squared when x is less than or equal to two or mx plus b when x is greater than two

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Answers: m = 4 and b = -4====== $f(x) = \begin{cases} x^2, & x \le 2 \\ mx + b,& x \ \textgreater \ 2 \end{cases}$ for f to be differentiable, we need the function to be continuous and the limit $\lim_{h \to 0}\big[ ( f(2+h) - f(2) ) / h \big]$ has to exist (the left-hand derivative is equal to the right-hand derivative)we see the pieces break at x = 2. the pieces themselves are continuous everywhere else. so we just need to establish that we need f to be continuous at x = 2. so this limit must exist for f to be continuous at x = 2: $f(2) = \displaystyle\lim_{x \to 2}f(x)$ the value of f(2) is 4 since we take the x² piece to evaluate it.for $\lim_{x \to 2}f(x)$ to exist, the left- and right-hand limits must be equal to each other $\displaystyle\lim_{x \to 2^-}f(x) = \lim_{x \to 2^+}f(x)$ left-hand limit we use f(x) = x² since that's for x < 2. the right-hand limit we use mx + b since that's for x > 2 $\displaystyle\lim_{x \to 2^-}x^2 = \lim_{x \to 2^+}(mx+b)$ by direct substitution $4 = 2m+b$ we do not have enough information right now to solve this. so now we examine the derivative of.the left-hand derivative of f at x = 2 we get by taking the piece x² (since that's for x < 2) and differentiating it: f'(x) = 2x when x < 2the right-hand derivative of f at x = 2 we get by taking the piece mx + b (since that's for x < 2) and differentiating it f'(x) = m when x > 2for this function to be "differentiable," it must be differentiable at every single point in its domain. x = 2 is in the domain of f so we need to make sure the derivative exists therethis limit must exist $\displaystyle \lim_{h \to 0} \frac{f(2+h) - f(2)}{h}$ for f'(x) to be differentiable at x = 2.for that limit to exist, the left- and right-hand limits must be equal to each other. $\displaystyle \lim_{h \to 0^-} \frac{f(2+h) - f(2)}{h} = \lim_{h \to 0^+} \frac{f(2+h) - f(2)}{h}$ the left-hand limit is the left-hand derivative at x = 2. the left-hand derivative at x = 2 is f'(2) = 2(2) = 4. the right-hand limit is the right-hand derivative at x=2, which is just m. $2(2) = m \implies m = 4$ so m has to be equal to 2.use this in the equation 4 = 2m+bwe obtained earlier to solve for b: 4 = 2(4) + b b = 4 - 8 b = -4m = 4, b = -4

answered: Guest

Iwould say false, not positive but if prices went up for both stores, technically there will be less buyers?

answered: Guest

Yea they are simplyfiy the ratios
Answer #10 i’ll give 10 points and mark you !

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Find the values of m and b that make the following function differentiable. the piecewise function f...