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Assume f(3)=1 and f ' (3) = 4 Let g(x)=x^2 + f(x) Find an equation of the line tangent to y=g(x) at x=3
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g(x)=x^2+f(x) g(3)=3^2+f(3)=10 g'(x)=d(x^2)/dx + f'(x) [ sum rule] =2x+f'(x) g'(3)=2(3)+4=10 So can you find the tangent line at (3,g(3))=(3,10) ?
yes, m=10 and I will use the point (3,1) correct?
since g(3)=10 (also), the point you need is (3,10)
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ok I can find it
y-10=10(x-3) y-10=10x-30 y=10x-20 Tangent line
Am I correct?
Yep!
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