Medium
Find the derivative of the function $f(x)=2x^{2}+3x-5$ at $x=-1$. Also,prove that $f^{\prime}(0)+3f^{\prime}(-1)=0$.
Given $f(x)=2x^{2}+3x-5$. The derivative $f^{\prime}(x)$ is given by $\frac{d}{dx}(2x^{2}+3x-5) = 4x+3$.
First,we find $f^{\prime}(-1)$:
$f^{\prime}(-1) = 4(-1)+3 = -4+3 = -1$.
Next,we find $f^{\prime}(0)$:
$f^{\prime}(0) = 4(0)+3 = 3$.
Now,we verify the expression $f^{\prime}(0)+3f^{\prime}(-1)$:
$f^{\prime}(0)+3f^{\prime}(-1) = 3 + 3(-1) = 3 - 3 = 0$.
Thus,it is proved that $f^{\prime}(0)+3f^{\prime}(-1)=0$.
First,we find $f^{\prime}(-1)$:
$f^{\prime}(-1) = 4(-1)+3 = -4+3 = -1$.
Next,we find $f^{\prime}(0)$:
$f^{\prime}(0) = 4(0)+3 = 3$.
Now,we verify the expression $f^{\prime}(0)+3f^{\prime}(-1)$:
$f^{\prime}(0)+3f^{\prime}(-1) = 3 + 3(-1) = 3 - 3 = 0$.
Thus,it is proved that $f^{\prime}(0)+3f^{\prime}(-1)=0$.
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