Respuesta :

LammettHash

The definition used here is

[tex]f'(c)=\displaystyle\lim_{x\to c}\dfrac{f(x)-f(c)}{x-c}[/tex]

so that

[tex]f'(-5)=\displaystyle\lim_{x\to-5}\dfrac{(2x^2+4x+7)-37}{x+5}=\lim_{x\to-5}\frac{2x^2+4x-30}{x+5}[/tex]

To compute the limit, factorize the numerator to get

[tex]2x^2+4x-30=2(x^2+2x-15)=2(x+5)(x-3)[/tex]

Then in the limit expression, the factors [tex]x+5[/tex] in the numerator and denominator cancel to give

[tex]f'(-5)=\displaystyle\lim_{x\to-5}2(x-3)=2(-5-3)=\boxed{-16}[/tex]

yoodyannapolis

f'(-5)=-16

We have  the formula :

[tex]f'(a)= \lim_{x \to {a}}\ \frac{f(x)-f(a)}{x-a}\\[/tex]

Here [tex]f(x) = 2x^2 + 4x + 7[/tex]

Substitute the value in the formula, we get

[tex]f'(a)= \lim_{x \to {-5}}\ \frac{(2x^2+4x+7)-(2(-5)^2+4(-5)+7)}{x+5}\\= \lim_{x \to {-5}}\ \frac{(2x^2+4x+7)-(50-20+7)}{x+5}\\= \lim_{x \to {-5}}\ \frac{(2x^2+4x-30)}{x+5}\\= \lim_{x \to {-5}}\ \frac{2(x+5)(x-3)}{x+5}\\=4(-5)+4\\=-16[/tex]

Therefore, f'(-5)=-16

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