Given a function, $$F(x)= \int_{-2}^x 3-4t \;\mathrm{dt}$$ find the equation of the line tangent to $F(x)$ at $x=1$
I'm having difficulty understanding why evaluating $F(1)$ (equal to $15$) is needed to supply the y-value of the tangent line. the tangent line apparently passes through $(1, 15)$ with a slope of $-1$, producing the equation $x + y = 16$
the difficulty arises in interpreting what is happening visually at $x = 1$. would graphing the antiderivative of $3-4t$ ($3t-2t^2+c$) be of any help?
Best Answer
The equation of the tangent line to the curve of $F$ at $x=1$ is
$$y=F(1)+F'(1)(x-1)$$ and $$F(1)=\int_{-2}^1(3-4t)dt=(3t-2t^2)\Bigg|_{-2}^1=15$$ and $$F'(1)=(3-4x)\Bigg|_{x=1}=-1$$ so the equation is
$$y=-x+16$$