Question

Find the point on the graph of the given function at which the slope of the tangent line is the given slope.f(x) = x³ + 12x² +57x+14; slope of the tangent line = 9The point at which the slope of the tangent line is 9 is(Simplify your answer. Type an ordered pair.)С2↑↑↑+BO-..XVVIV

Answer 1

We have a function:

`f(x)=x^3+12x^2+57x+14`

We have to find the point at which the slope of the tangent line is m = 9.

The slope of the tangent line of a function at a point is given by the value of the first derivative at that point, so we start by finding the first derivative:

`\begin{gathered} f^(\prime)(x)=3x^2+12(2x)+57(1)+14(0) \\ f^(\prime)(x)=3x^2+24x+57 \end{gathered}`

We have now to find the value of x for which this derivative is equal to 9, so we can write:

`\begin{gathered} f^(\prime)(x)=9 \\ 3x^2+24x+57=9 \\ 3(x^2+8x+19)=3(3) \\ x^2+8x+19=3 \\ x^2+8x+19-3=0 \\ x^2+8x+16=0 \end{gathered}`

We now have to find the roots of this equation to find the solution for x:

`\begin{gathered} x=(-8\pm√(8^2-4(1)(16)))/(2(1)) \\ x=(-8\pm√(64-64))/(2) \\ x=(-8\pm√(0))/(2) \\ x=-4 \end{gathered}`

This slope for the tangent line only happens for the point located at x = -4.

We can find the y-coordinate as y = f(-4):

`\begin{gathered} f(-4)=(-4)^3+12(-4)^2+57(-4)+14 \\ f(-4)=-64+12(16)-228+14 \\ f(-4)=-64+192-228+14 \\ f(-4)=-86 \end{gathered}`

We can check this solution with a graph as:

Answer:

(-4, -86)