Find the equation of the line tangent to the graph of the given function at the point with the indicated x-coordinate.
f(x) = (x^0.7 + 1)(x^2 + x); x=1

Question

Find the equation of the line tangent to the graph of the given function at the point with the indicated x-coordinate.
f(x) = (x^0.7 + 1)(x^2 + x);  x=1

jhannybean · Answer

There isn't even a point given? just x=1? :\

jhannybean · Answer

well, first expand f(x) so you can take the derivative..$$f(x)=(x^{0.7}+1)(x^2+x)$$$$f(x) = x^{2.7}+x^{1.7}+x^2+x$$take the derivative..$$f'(x)=2.7x^{1.7}+1.7x^{0.7}+2x+1$$ plug in x=1$$f'(x)=2.7(1)^{1.7}+1.7(1)^{0.7}+2(1)+1$$$$f'(x)=7.4$$ now you can plug in the slope, 7.4 and x=1 into $$y=mx+b$$ to solve for the equation of the tangent line.

jhannybean · Answer

sorry, your $$f(x)=x^{7.4}$$ and then your derivative would be $$f'(x)=7.4 x^{6.4}$$

jhannybean · Answer

but if you were to plug in x=1... your slope would still = 7.4...

anonymous · Answer

How do I know what y is when I use the formula to find the tangent line?

jhannybean · Answer

you have $$y=mx+b$$ and you're given x=1 and found your slope = 7.4, but you don't have a point which you could use to plug into y....$$y=7.4(1)+b$$ that doesn't make sense to me :\ no point given...

anonymous · Answer

@jhannybean, the "point" would be $(x,f(x))$. Since $x=1$, the point you would use is $(1,f(1))=(1,4)$.

jhannybean · Answer

ohhh... thank you!!! :D i've forgotten. haha.

jhannybean · Answer

okay,so you've found your m= 7.4and the point (1,4) into y=mx+b

4=(7.4)(1)+b
-7.4+4 =b
-3.4=b

so reform your slope intercept form equation. 

y=mx+b
y=7.4x -3.4

jhannybean · Answer

Thank you @SithsAndGiggles I was stumped fora minute. haha.

anonymous · Answer

Ok. Thank you very much!!

jhannybean · Answer

no problem :)

anonymous · Answer

@Jhannybean , you're welcome