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?
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$\begingroup$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
$$y=-x+16$$
$\endgroup$ $\begingroup$Some visualization: With this definition of $F(x)$, direct integration gives $F(x) = \left[3t-2t^2\right]_{-2}^x =14+ 3x-2x^2$. This is easily plotted in WolframAlpha along with the tangent line $y=16-x$ you found.
To interpret this, though, it helps to plot $F(x)$ along with $F'(x)=3-2t$ (WolframAlpha). Observe that $F'(x)$ is positive up until $x=3/4$ and therefore the area under $F'(x)$ (i.e. $F(x)$) is increasing. But past $x=3/4$ $F'(x)$ becomes negative and so the area begins to decrease. Consequently the slope of $F(x)$ at $x=1>3/4$ is negative.
$\endgroup$ 1 $\begingroup$You have $y=F(x)$. The equation of the tangent line is $$y-y(1) = y'(1)(x-1).$$ Since $y'(1) F'(1) = 3-4(1) = -1$ and $y(1) = F(1) = 15$ it follows $$y-15=(-1)(x-1) \text{ and } x+y=16.$$
$\endgroup$ $\begingroup$Above is a graph of a function (blue curve) and a set of parallel red lines. The bottom two red lines cut the blue curve, the top five red lines don't even touch it. Only one red line intersects the blue curve (locally) at exactly one point, even though every red line has the same slope as the blue line at that point's horizontal component. You can easily see why you need to know the slope, as well as the coordinates of the point of tangency to uniquify the tangent line.
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