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The equation respenting the left table is

Does the function satisfy the hypotheses of the mean value theorem on the given interval? f(x) = 4x^2 + 3x + 4, [−1, 1] no, f is continuous on [−1, 1] but not differentiable on (−1, 1). no, f is not continuous on [−1, 1]. yes, f is continuous on [−1, 1] and differentiable on (−1, 1) since polynomials are continuous and differentiable on . there is not enough information to verify if this function satisfies the mean value theorem. yes, it does not matter if f is continuous or differentiable; every function satisfies the mean value theorem.

For [tex]f(x) = 4x + 1[/tex] and (x) = [tex]g(x)= x^{2} -5,[/tex] find [tex](\frac{g}{f}) (x)[/tex]a. [tex]\frac{x^{2} - 5 }{4x +1 },x[/tex] ≠ [tex]-\frac{1}{4}[/tex]b. x[tex]\frac{4 x +1 }{x^{2} - 5}, x[/tex] ≠ ± [tex]\sqrt[]{5}[/tex]c. [tex]\frac{4x +1}{x^{2} -5}[/tex]d.[tex]\frac{x^{2} -5 }{4x + 1}[/tex]