Question

Find the derivative of the function $f\left(x\right)=2x^2+3x-5$ at $x=-1$. Also prove that $f^{\prime }\left(0\right)+3f^{\prime }(-1)=0$

Answer 1

Step $1:$
Given $f\left(x\right)=2x^2+3x-5$
We know that $f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\left(2\right(x+h{)}^2+3(x+h)-5)-(2x^2+3x-5)}{h}$
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\left(2\right(x^2+h^2+2xh)+3(x+h)-5)-(2x^2+3x-5)}{h}$
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{2x^2+2h^2+4xh+3x+3h-5-2x^2-3x+5}{h}$
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{2h^2+4xh+3h}{h}$
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{h(2h+4x+3)}{h}$
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}(2h+4x+3)$
$\Rightarrow f^{\prime }\left(x\right)=4x+3$

Putting $x=-1$
$\Rightarrow f^{\prime }(-1)=4(-1)+3$
$\Rightarrow f^{\prime }(-1)=-4+3$
$\Rightarrow f^{\prime }(-1)=-1$
Hence, $f^{\prime }(-1)=-1$

Step 2: Solve for proving
For $f^{\prime }\left(0\right)$
$\Rightarrow f^{\prime }\left(x\right)=4x+3$
$\Rightarrow f^{\prime }\left(0\right)=4\cdot 0+3$
$\Rightarrow f^{\prime }\left(0\right)=3$
Now,
$f^{\prime }\left(0\right)+3f^{\prime }(-1)=3+3(-1)$
$=0$
Hence proved.
Therefore, value of $f^{\prime }(-1)$ is $-1$ and its proved that $f^{\prime }\left(0\right)+3f^{\prime }(-1)=0$.