(1 point) Find a and b so that the function f(x)= ) 5x^3-6x^2+8,&xlt -2 ax+b,&xgeqslant -2 is both continuous and differentiable. a= b=

To ensure that the function $f(x)$ is both continuous and differentiable at $x = -2$, we need to make sure that the left-hand limit and the right-hand limit are equal at $x = -2$, and that the derivatives from both sides are also equal at that point.<br /><br />First, let's find the left-hand limit as $x$ approaches $-2$:<br />$\lim_{{x \to -2^-}} f(x) = \lim_{{x \to -2^-}} (5x^3 - 6x^2 + 8) = 5(-2)^3 - 6(-2)^2 + 8 = -40 - 24 + 8 = -56$<br /><br />Next, let's find the right-hand limit as $x$ approaches $-2$:<br />$\lim_{{x \to -2^+}} f(x) = \lim_{{x \to -2^+}} (ax + b) = a(-2) + b = -2a + b$<br /><br />For the function to be continuous at $x = -2$, the left-hand limit and the right-hand limit must be equal:<br />$-56 = -2a + b$<br /><br />Now, let's find the derivative of the left-hand side of the function:<br />$f'(x) = \frac{d}{dx}(5x^3 - 6x^2 + 8) = 15x^2 - 12x$<br /><br />Next, let's find the derivative of the right-hand side of the function:<br />$f'(x) = \frac{d}{dx}(ax + b) = a$<br /><br />For the function to be differentiable at $x = -2$, the derivatives from both sides must be equal:<br />$15(-2)^2 - 12(-2) = a$<br />$60 + 24 = a$<br />$a = 84$<br /><br />Now, substitute $a = 84$ into the equation $-56 = -2a + b$ to find $b$:<br />$-56 = -2(84) + b$<br />$-56 = -168 + b$<br />$b = -56 + 168$<br />$b = 112$<br /><br />Therefore, the values of $a$ and $b$ that make the function $f(x)$ both continuous and differentiable are:<br />$a = 84$<br />$b = 112$