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Question

Find a possible function $h (x)$ and a number $c$ such that the following limit represents the slope of the curve $y = h (x)$ at $(c, h (c))$ . Then, evaluate the limit:

$\lim_{x \to - 1} \frac{4 x^{2} - 2 x - 6}{x + 1}$

Accepted Answer

$h (x) = 4 x^{2} - 2 x$

$c = - 1$

$\lim_{x \to - 1} \frac{4 x^{2} - 2 x - 6}{x + 1}$ $=$ $negative 10$

Answer

$h (x) = 4 x^{2} - 2 x$

$c = - 1$

$\lim_{x \to - 1} \frac{4 x^{2} - 2 x - 6}{x + 1}$ $=$ $negative 10$

Answer

$h (x) = 4 x^{2} - 2 x$

$c = - 1$

$\lim_{x \to - 1} \frac{4 x^{2} - 2 x - 6}{x + 1}$ $=$ $negative 10$

Answer

$h (x) = 4 x^{2} - 2 x$

$c = - 1$

$\lim_{x \to - 1} \frac{4 x^{2} - 2 x - 6}{x + 1}$ $=$ $negative 10$

Find a possible function h(x)h\left(x\right) and a number cc such that the following limit represents the slope of the curve y=h(x)y=h\left(x\right) at (c,h(c))(c,h(c)). Then, evaluate the limit:limx→−14x2−2x−6x+1{\displaystyle\lim_{x\to-1}}\frac{4x^2-2x-6}{x+1}

Find a possible function $h (x)$ and a number $c$ such that the following limit represents the slope of the curve $y = h (x)$ at $(c, h (c))$ . Then, evaluate the limit:
$\lim_{x \to - 1} \frac{4 x^{2} - 2 x - 6}{x + 1}$