Question

$Find \: the \: tangent \: line \: \\ to \: f(x) = {7x}^{4} + {8x}^{ - 6} + 2x \: at \: x = 1.$
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Answer 1

Answer:

y=22×+35 is the tangent line

given expression is F (x)=7×^4+8/x^6+2x

Differentiating given equation F'(x) gives 28x^3+

-48x^-7+2, i.e -28-48/('1)^7+2=-29+48+2=22=slope.

F(x)=y=7(-1)^4+8(-1)^-6+2(-1)=7+8/(-1)^6-2=7+8-2=13 is y co-ordinate at the point.

equation of tangent y=mx+c and at point x, y=-1,13 with m=22, we obtain c by substituting the values of x,y,m into y=mx+c to give

13=22 (-1)+(or 23+22)=(=35.

y=22+35 is the equation fir tangent

if 8×-6 was intended of 8x-6, then

Answer =-18×-27

Given x=-1, so (x)=F(-1)=7(-1)^4+10*(-1)-6=7-10-6=-9=y

co-ordinate.

Differentiating F (x)=F'(x)=7*(4×^3)+10=28×^3+10

Slope at given co-ordinate -1=F(-1)=2(-1)*3+10=-28+

10=-18=m

This y=-9, m=slope=-18,x=-1 at the point

Approach A) using slop - point notation, we get y-y1=m)x-x1)

=y-(-9)=-18(x-(-1))

so y+9=-18 (x+1) or y=-18×-27 is the equation of tangent at -1

Approach B) as we know slope=m=-18, we know can write the lines equation as

y=mx+c, but we know x,y and m which gives

-9=(-18)(-1)+c or

-9=18+( so c =-27 or

y=mx+c=-18×-27 at the point x=1

Step-by-step explanation:

I HOPE IT HELPFUL YOU