Determine f'(x) from first principles if f(x)=$-5x^{2\ }+\ x$ . Hence calculate the tangent to f(x) where x=1.

Question

So what I did to find the answer to the first part of the question is: f'(x)=$-10x+1$

I worked this out fine but when I got to the second part, it got a bit confusing. This is what I did: Substituted $x=1$

$y=-5(1)^2+1$

and got $y= -4$

Then used the equation of the tangent: $y-y_1=m\left(x-x_1\right)$ which leads to $y+4=-5\left(x-1\right)$ $y=-5x+1$

But the memo says this is incorrect. Can someone please explain why?

Answer 1

We are asked to find the equation of tangent to the curve $f(x) = -5x^2+x$ at $x = 1$.

When $x=1$ then value of $y$ will be,

$\implies f(1) = -5(1)^2 + (1)\\ \implies f(1) = -5 + 1 = -4$

Therefore, we got a point $(1,-4)$ from where the line passes.

Slope of the tangent at $x=1$ is given by $f'(1) = -9$.

So, using point slope form:-

$\implies (y-(-4)) = -9(x-1)\\$

$\implies (y+4) = -9x+9\\$

$\implies y = -9x+9-4\\$

$\implies y = -9x+5\\$

Therefore the required equation is $y = -9x + 5$