Given the function f(x) = 4x3 + 13x, what is the gradient of the function when x = 2 ?

y=45x+115

Step-by-step explanation:

We can find the slope of a line tangent to a curve at a point by evaluating the derivative of the function at that point.

We are given the function

f ( x ) = 4 x 3 + 12 x 2 + 9 x + 7

= f ( x ) = 4 x 3 + 12 x 2 + 9 x 1 + 7 x 0

Using the power rule, let's now compute the derivative of  f( x )

f ' ( x ) = ( 3 ⋅ 4 x 2 ) + ( 2 ⋅ 12 x ) + ( 1 ⋅ 9 x 0 ) + 0 ⋅ 7 x − 1

f ' ( x ) = 12 x 2 + 24 x + 9

We can now find the slope of  f ( x )  at  x = − 3  by substituting this value into f ' ( x )

f ' ( − 3 ) = 12 ( − 3 ) 2 + 24 ( − 3 ) + 9

f'(-3)=12(9)+24(-3)+9

f'(-3)=108-72+9

f'(-3)=45(slope of the tangent line at  x = − 3

Now that we have a slope for the tangent line, we need to identify a point on the line.

We know the tangent line touches the function  f ( x )  at the point  x = − 3 , so let's find the value of  f ( x )  at this point:

f(-3)=4(-3)3+12(-3)2+9(-3)+7

f(-3)=4(-27)+12(9)+9(-3)+7

f(-3)=108+108-27+7

f'(-3)=-20

So we know the tangent line goes through the point

(-3,-20)

Finally, we can use the point-slope formula for a line to find the equation of the tangent line.

y=mx+b

To find the value of  b , substitute the values we have calculated for the point and slope of the tangent line:

(-20)=(45)(-3)+b

-20=-135+b

b=115

So our final answer for the equation of the tangent line is:

y=45x+115