given is F(x)= the integral from x=1 to x=2x^3 of the function (3t-4)^2 dt
I need to find the equation of the tangent line to F(x) at x=-1
so far i have:
f=600 (which is the slope of F at x=-1)
i need help finding the (x,y) coordinates
given is F(x)= the integral from x=1 to x=2x^3 of the function (3t-4)^2 dt
I need to find the equation of the tangent line to F(x) at x=-1
so far i have:
f=600 (which is the slope of F at x=-1)
i need help finding the (x,y) coordinates
$\displaystyle F(x)=\int_1^{2x^3}(3t-4)^2dt$
Let $\displaystyle u=2x^3$
$\displaystyle F'(x)=\frac{du}{dx}\frac{d}{du}\int_1^u(3t-4)^2dt$
$\displaystyle F'(x)=6x^2[3(2x^3)-4]^2$
$\displaystyle F'(-1)=600$
Ok, In get the same thing there. Think of the equation of the tangent line as follows:
$\displaystyle y-F(-1)=600[x-(-1)]$
$\displaystyle y-F(-1)=600(x+1)$
So you need to find
$\displaystyle F(-1)=\int_1^{2(-1)^3}(3t-4)^4dt$
So that's just a basic definite integral right? So just evaluate F(-1) by using the substitution rule.